CBSE Class 12 Physics Electrostatics Boards Questions Worksheet

Unit I: Electrostatics

This unit introduces the study of electric charges at rest, exploring their fundamental properties, interactions, and fields. Key topics include Chapter 1 on Electric Charges and Fields (Coulomb's Law, field calculations, dipole behavior, Gauss's Theorem) and Chapter 2 on Electrostatic Potential and Capacitance (potential differences, equipotential surfaces, dielectric polarization, and capacitive network analysis).

Question 101. State Coulomb’s law in electrostatics.
Answer: According to Coulomb's law, the force of attraction or repulsion acting between any two stationary electric point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance separating them.
Mathematically, if \( q_1 \) and \( q_2 \) are two point charges separated by a distance \( r \) in a vacuum: \[ F \propto \frac{q_1 q_2}{r^2} \] This can be written as: \[ F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} \] where \( \frac{1}{4 \pi \varepsilon_0} \) is the electrostatic proportionality constant.

In simple words: Coulomb's law explains how strongly two stationary electric charges push or pull on each other. This force gets stronger if the charges are larger, and weaker if they are farther apart.

Exam Tip: Draw the simple diagram of two point charges \( q_1 \) and \( q_2 \) separated by distance \( r \) to secure full marks.

Question 102. Write Coulomb’s law in vector form. What is the importance of expressing it in vector form ?
Answer: Coulomb's law in vector form is written as: \[ \vec{F}_{12} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \hat{r}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{|\vec{r}|^3} \vec{r}_{21} \] where \( \vec{F}_{12} \) is the force exerted on charge \( q_1 \) by charge \( q_2 \), and \( \hat{r}_{21} \) is the unit vector pointing from \( q_2 \) to \( q_1 \).

**Importance of the Vector Form:**
(i) Since \( \hat{r}_{12} = -\hat{r}_{21} \), it follows that \( \vec{F}_{12} = -\vec{F}_{21} \). This proves that the electrostatic forces between two charges are equal and opposite, thereby obeying Newton's third law of motion.
(ii) Because the electrostatic force acts along the line connecting the centers of the two charges, it is classified as a central force.
In simple words: Writing Coulomb's law with vectors shows the exact direction of the force. It proves that when charge A pushes charge B, charge B pushes back on A with the same strength in the opposite direction.

Exam Tip: Clearly define the unit vectors like \( \hat{r}_{12} \) and state how they relate to the direction of attraction or repulsion.

Question 103. Write any two limitations of Coulomb’s law.
Answer: Two major limitations of Coulomb's law are:
(i) It is applicable only to stationary point charges; it fails when charges are in continuous motion or have complex, extended geometries.
(ii) It ceases to be valid for extremely short subatomic distances less than \( 10^{-15}\text{ m} \) (the scale of atomic nuclei) where strong nuclear forces dominate.
In simple words: Coulomb's law only works when charges are perfectly still and tiny like points, and it stops working at extremely small scales inside an atomic nucleus.

Exam Tip: Mention "point charges" and "nuclear distances" as the two primary boundary conditions.

Question 104. (a) Name any two basic properties of electric charge.
(b) What does q1 + q2 = 0 signify in electrostatics ?
Answer: (a) Two fundamental characteristics of electric charge are:
(i) **Quantization of Charge:** Electric charge exists in discrete packets of integral multiples of the elementary charge \( e \).
(ii) **Conservation of Charge:** The total electric charge of an isolated system remains constant over time.

(b) The equation \( q_1 + q_2 = 0 \implies q_1 = -q_2 \) indicates that electric charges are additive algebraically, and here, the two charges are equal in magnitude but opposite in polarity (sign).
In simple words: (a) Charge can only exist in set packets and cannot be created or destroyed. (b) Adding the charges to get zero means they are identical in size but one is positive and the other is negative.

Exam Tip: Explain quantization mathematically using \( q = \pm ne \) to secure complete marks.

Question 105. Is the force acting between two point electric charges q1 and q2 kept at some distance apart in air, attractive or repulsive when (a) q1 q2 > 0 (b) q1 q2 < 0 ? 
Answer: The behavior of the force depends on the product of the charges:
(a) When \( q_1 q_2 > 0 \), both charges have the same sign (either both positive or both negative). Thus, the force acting between them is **repulsive**.
(b) When \( q_1 q_2 < 0 \), the charges have opposite signs (one positive, one negative). Thus, the force acting between them is **attractive**.
In simple words: If the product of two charges is positive, they repel each other. If it's negative, they pull together because opposites attract.

Exam Tip: Connect this directly to the fundamental law of electrostatics: "like charges repel, unlike charges attract."

Question 106. Two insulated charged copper spheres A and B of identical size have charges qa and -3qa respectively. When they are brought in contact with each other and then separated, what are the new charges on them ? 
Answer: When two identical conducting spheres are brought into contact, the total charge distributes itself equally between them due to electrostatic conduction.
The total charge of the system is: \[ q_{\text{total}} = q_a + (-3q_a) = -2q_a \] Upon separating them, this charge divides equally: \[ q_{\text{new}} = \frac{q_{\text{total}}}{2} = \frac{-2q_a}{2} = -q_a \] Therefore, each sphere will have a new charge of \( -q_a \).
In simple words: When you touch identical charged balls, they share their total charge equally. Here, the net charge of minus two is shared, leaving each ball with minus one charge.

Exam Tip: Emphasize that this equal sharing of charge occurs only because the two spheres are identical in size.

Question 107. Define dielectric constant of a medium in terms of force between electric charges. What is its S.I. unit ? 
Answer: The dielectric constant (\( K \)) of a given medium is defined as the ratio of the electrostatic force (\( F_{\text{vacuum}} \)) between two point charges separated by a certain distance in a vacuum to the electrostatic force (\( F_{\text{medium}} \)) between the same two charges kept at the same distance in that medium: \[ K = \frac{F_{\text{vacuum}}}{F_{\text{medium}}} \] Since it is the ratio of two forces, the dielectric constant is a dimensionless quantity and has **no SI unit**.
In simple words: The dielectric constant is a number that tells you how many times weaker the electrical force between two charges becomes when you put them inside a material instead of empty space.

Exam Tip: Explicitly write "no unit" or "dimensionless" to avoid losing marks on the unit question.

Question 108. How does the Coulomb force between two point charges depend upon the dielectric constant of the intervening medium ? 
Answer: The electrostatic Coulomb force \( F \) acting between two point charges is inversely proportional to the dielectric constant \( K \) of the medium separating them: \[ F = \frac{1}{4 \pi \varepsilon_0 K} \frac{q_1 q_2}{r^2} \implies F \propto \frac{1}{K} \] Thus, introducing a medium with a higher dielectric constant reduces the force between the charges.
In simple words: The electrical force between two charges decreases when they are in a medium with a larger dielectric constant because the medium shields the charges.

Exam Tip: State the inverse proportionality relation clearly in mathematical terms.

Question 109. Two same balls having equal positive charge Coulombs are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two ?
Answer: The electrostatic repulsive force acting between the two charged balls is given by: \[ F = \frac{1}{4 \pi \varepsilon_0 K} \frac{q^2}{r^2} \] When a plastic sheet (which acts as a dielectric medium with a dielectric constant \( K > 1 \)) is inserted between the balls, the force between them decreases because the force is inversely proportional to the dielectric constant (\( F \propto \frac{1}{K} \)).
In simple words: Inserting a plastic sheet between the suspended charged balls weakens the electrical force pushing them apart because plastic reduces the transmission of electric fields.

Exam Tip: Explicitly mention that \( K > 1 \) for plastic to justify why the force must decrease.

Question 110. Force between two point electric charges kept at a distance d apart in air is F. If the charges are kept at the same distance in water, how does the force between them change ? 
Answer: The electrostatic force in a medium of dielectric constant \( K \) is related to the force in air \( F \) by: \[ F_{\text{water}} = \frac{F}{K} \] Since the dielectric constant of water is \( K \approx 80 \), the electrostatic force between the two charges when placed in water decreases to \( \frac{F}{80} \) of its original value in air.
In simple words: Putting the charges in water weakens the force between them by about 80 times, because water has a very high dielectric constant.

Exam Tip: Mention the approximate value of the dielectric constant of water (\( K = 80 \)) to make your numerical estimation precise.

Question 111. Two point charges having equal charges separated by distance experience a force of 8N. What will be the force experienced by them, if they are held in water, at the same distance ? (Given : K_water = 80 ) 
Answer: The electrostatic force experienced by charges in water is given by: \[ F_{\text{water}} = \frac{F_{\text{air}}}{K_{\text{water}}} \] Substituting the given values, where \( F_{\text{air}} = 8\text{ N} \) and \( K_{\text{water}} = 80 \): \[ F_{\text{water}} = \frac{8}{80} = 0.1\text{ N} \] The new force experienced by the charges in water is \( 0.1\text{ N} \).
In simple words: In water, the original force of 8 Newtons gets divided by the dielectric constant of 80, dropping to just 0.1 Newtons.

Exam Tip: Show the substitution clearly to gain full marks for this simple numerical.

Question 112. Does the charge given to a metallic sphere depend on whether it is hollow or solid ? Give reason for your answer. 
Answer: No, the maximum charge that can reside on a metallic sphere does not depend on whether it is hollow or solid.
**Reason:** Any excess charge given to a conductor distributes itself and resides solely on its outer surface to minimize potential energy. Since both solid and hollow spheres of equal radius have identical outer surface areas, they can hold the same maximum charge.
In simple words: No, it doesn't matter. Since all excess charge resides only on the outside surface of a metal ball, solid and hollow spheres of the same size hold identical amounts of charge.

Exam Tip: State clearly that "charge resides only on the outer surface of a conductor" as the core physics principle.

Question 113. A comb run through one’s dry hair attracts small bits of paper. Why ? What happens if the hair is wet or if it is a rainy day ? 
Answer: 1. When run through dry hair, the comb becomes electrostaticly charged due to friction. When brought near bits of paper, it polarizes the molecules in the paper, creating an attractive electrostatic force that lifts them.
2. If the hair is wet or if it is a humid, rainy day, the moisture acts as a conducting path that neutralizes the charges. This drastically reduces friction, preventing the comb from getting charged and thus preventing it from attracting paper.
In simple words: Rubbing a comb on dry hair charges it up, letting it attract paper by polarizing it. On a wet or rainy day, water conducts the charge away before the comb can build up any static power.

Exam Tip: Use the key term "polarization of molecules" to explain why the uncharged paper is attracted to the charged comb.

Question 114. Define electric field intensity. Write its S.I. unit. Is it a scalar or vector quantity ? 
Answer: Electric field intensity at any point is defined as the electrostatic force experienced per unit positive test charge placed at that point, provided the test charge is infinitesimally small so as not to disturb the source charge configuration: \[ \vec{E} = \lim_{q_0 \to 0} \frac{\vec{F}}{q_0} \]
* **SI Unit:** Newton per Coulomb (\( \text{N/C} \)) or Volt per meter (\( \text{V/m} \)).
* **Type of quantity:** It is a **vector** quantity, directed along the force experienced by a positive test charge.
In simple words: Electric field intensity measures how hard an electric field pushes on a tiny positive charge. It has a specific direction and is measured in Newtons per Coulomb.

Exam Tip: Write down the limit notation \( \lim_{q_0 \to 0} \) to demonstrate rigorous conceptual understanding.

Question 115. The electric field intensity at any point is defined as \( \lim_{q_0 \to 0} \frac{F}{q_0} \). What is the physical significance of the term \( \lim_{q_0 \to 0} \) in this expression ?
Answer: The condition \( \lim_{q_0 \to 0} \) signifies that the magnitude of the test charge \( q_0 \) must be infinitesimally small. This ensures that the test charge does not generate its own significant electric field, which would otherwise alter the spatial distribution of the source charges and modify the very electric field being measured.
In simple words: The limit ensures that our test charge is too small to push around the main charges we are trying to measure, keeping the field reading accurate.

Exam Tip: Focus on the phrase "does not alter the distribution of source charges."

Question 116. (i) What is the physical significance of electric field ?
(ii) Write an expression for force acting on a test charge q_0 placed in a uniform electric field.
Answer: (i) **Physical Significance:** The concept of an electric field provides a way to explain "action at a distance" without direct contact. It describes how any charge alters the surrounding space, so that any other charge placed in this space immediately experiences a local electrostatic force.
(ii) **Force Expression:** The electrostatic force \( \vec{F} \) acting on a test charge \( q_0 \) is: \[ \vec{F} = q_0 \vec{E} \]
In simple words: (i) The electric field acts as an invisible force field that charges cast around themselves. (ii) The push on any charge is just the charge value multiplied by the field strength.

Exam Tip: Write the force equation in vector notation to show the directional alignment of \( \vec{F} \) and \( \vec{E} \).

Question 117. A proton is placed in a uniform electric field directed along the positive x-axis. In which direction will it tend to move ? 
Answer: Since a proton carries a positive electric charge, the electrostatic force acting on it is directed parallel to the electric field vector. Therefore, the proton will accelerate along the **positive x-axis** (in the direction of the electric field).
In simple words: Positive charges are pushed in the same direction that the electric field points, so the proton moves along the positive x-axis.

Exam Tip: Contrast this with an electron, which would move in the opposite direction (along the negative x-axis).

Question 118. Why must electrostatic field at the surface of a charged conductor be normal to the surface at every point ? Give reason. 
Answer: Under electrostatic equilibrium conditions, the surface of a charged conductor behaves as an equipotential surface (\( V = \text{constant} \)).
The work done in moving a charge through a small displacement \( \vec{dr} \) along the surface is: \[ dV = -\vec{E} \cdot \vec{dr} \] Since the potential \( V \) is constant everywhere on the surface, \( dV = 0 \): \[ \vec{E} \cdot \vec{dr} = 0 \implies E \, dr \cos\theta = 0 \] Because \( E \neq 0 \) and \( dr \neq 0 \), it follows that: \[ \cos\theta = 0 \implies \theta = 90^\circ \] Therefore, the electric field \( \vec{E} \) must be perpendicular (normal) to the surface at every point. If it weren't, a tangential field component would cause surface currents, violating electrostatic equilibrium.
In simple words: If the electric field lines weren't perpendicular, there would be a sideways electric push that would make the electrons on the surface flow in currents. Since the charges are stationary, the field must point straight out.

Exam Tip: Use the mathematical proof \( \vec{E} \cdot \vec{dr} = 0 \implies \theta = 90^\circ \) to provide a rigorous, high-scoring answer.

Question 119. Define electric potential at a point. Write its S.I. unit. Is it potential a scalar or vector ? 
Answer: Electric potential at any point in an electric field is defined as the amount of work done by an external force in moving a unit positive charge from infinity to that point against the electrostatic forces of the field: \[ V = \frac{W_{\infty \to P}}{q_0} \]
* **SI Unit:** Joule per Coulomb (\( \text{J/C} \)) or Volt (\( \text{V} \)).
* **Type of quantity:** It is a **scalar** quantity.
In simple words: Electric potential is the energy needed to bring a unit of positive charge from far away to a specific point. It is measured in Volts and doesn't have a direction.

Exam Tip: Clearly state both "Volt" and "Joule/Coulomb" to demonstrate complete unit knowledge.

Question 120. Name the physical quantity whose S.I. unit is J/C. Is it a scalar or vector quantity ? 
Answer: The identity and nature of the quantity are:
* **Physical Quantity:** The physical quantity is **electric potential** (or electromotive force / potential difference).
* **Type of quantity:** It is a **scalar** quantity.
In simple words: Joules per Coulomb is the definition of a Volt, which is the unit for electric potential. It is a scalar quantity.

Exam Tip: Connecting the unit \( \text{J/C} \) directly to the definition of electric potential is the key to full marks.

Question 121. Why is the potential inside a hollow spherical charged conductor constant and has the same value as on its surface ? 
Answer: The electric field \( E \) and electric potential \( V \) are related by: \[ E = -\frac{dV}{dr} \implies dV = -E \, dr \] Inside a hollow charged conducting sphere, there are no charges enclosed, so the electric field is zero (\( E = 0 \)).
Substituting this: \[ dV = 0 \implies V = \text{constant} \] Since the change in potential is zero, the electric potential remains constant at all internal points and is equal to its value on the outer surface.
In simple words: Since there is no electric field inside a hollow metal shell, you don't need to do any work to move charges around inside. This means the voltage is completely flat and matches the surface voltage.

Exam Tip: State both \( E = 0 \) inside and \( E = -\frac{dV}{dr} \) to build a mathematically solid explanation.

Question 122. A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 5 V. What is the potential at the centre of the sphere ? 
Answer: The electric potential inside a hollow charged metal sphere is constant and equals the potential on its surface.
Therefore, since the surface potential is \( 5\text{ V} \), the potential at the center of the sphere is also **\( 5\text{ V} \)**.
In simple words: Because the voltage inside a hollow metal sphere is completely uniform, the voltage at the center is exactly the same as on the outside surface, which is 5 Volts.

Exam Tip: Do not waste time doing calculations with the radius; state the conceptual rule directly to get full marks.

Question 123. A point charge +Q is placed at a point O as shown in the figure. Is the potential difference VA - VB positive, negative or zero ? 
Answer: The electric potential \( V \) due to a point charge \( Q \) at a distance \( r \) is: \[ V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r} \implies V \propto \frac{1}{r} \] Since point A is closer to the positive charge than point B (\( r_A < r_B \)), its potential is higher: \[ V_A > V_B \] Therefore, the potential difference \( V_A - V_B \) is **positive**.

In simple words: Since point A is closer to the positive charge, it has a higher voltage than point B. Subtracting the smaller voltage from the larger one gives a positive result.

Exam Tip: Clearly state the inverse relationship \( V \propto \frac{1}{r} \) for a positive charge to justify your conclusion.

Question 124. Define electric line of force/electric field line.
Answer: An electric field line is defined as an imaginary continuous path (either straight or curved) drawn in an electric field such that the tangent to it at any point indicates the direction of the electric field vector at that point. It represents the path along which an isolated, free positive unit test charge would move.
In simple words: An electric field line is an imaginary path that shows which way a tiny positive charge would float if you let it loose in the electric field.

Exam Tip: Mention both "imaginary path" and the "tangent direction" to cover all grading criteria.

Question 125. State any two properties of electric field lines. 
Answer: Two primary properties of electric field lines are:
1. They originate from positive charges and terminate on negative charges, never forming closed continuous loops.
2. No two electric field lines can ever intersect each other, as this would imply two different directions of the electric field at the point of intersection.
In simple words: Field lines start on positive charges and end on negative ones without forming loops. Also, they never cross because a point can't have two field directions at once.

Exam Tip: State why they do not intersect, as that is a very common sub-question on exams.

Question 126. What is the importance of electric field lines ? 
Answer: The significance of electric field lines includes:
1. The tangent drawn at any point on a field line gives the direction of the electric field at that point.
2. The relative density (closeness) of the field lines in a region represents the relative strength of the electric field there; closer lines indicate a stronger field.
In simple words: They help us visualize the electric field: a tangent shows the field's direction, and crowded lines show where the field is strongest.

Exam Tip: Use the term "crowdedness/relative closeness" to describe field strength.

Question 127. Why do the electrostatic filed lines not form closed loops ? 
Answer: Electrostatic field lines do not form closed loops because the electrostatic field is conservative in nature. This means they must start at a source (positive charge) and end at a sink (negative charge), rather than looping back to their starting point.
In simple words: Because electrostatic forces are conservative, field lines always run from positive to negative charges and never loop back on themselves like magnetic lines do.

Exam Tip: Use the key phrase "conservative nature of the electrostatic field" to secure full marks.

Question 128. Why do the electric field lines never cross each other ? 
Answer: If two electric field lines were to cross, we could draw two different tangents at the single point of intersection. This would mean the electric field vector has two distinct directions at the same point, which is physically impossible.
In simple words: If they crossed, a charge placed at the intersection wouldn't know which way to go because the field would point in two directions at once, which can't happen.

Exam Tip: Mention that a vector quantity like electric field can only have one unique direction at any given point.

Question 129. Why do the electrostatic filed lines are always normal to the surface of a conductor 
Answer: If the electric field lines were not perpendicular to the conductor's surface, the field would have a non-zero tangential component along the surface. This component would exert forces on the free surface electrons, causing them to move and generate electric currents. Since the system is in electrostatic equilibrium (stationary charges), no such currents can exist, meaning the field must be purely normal.
In simple words: If the field lines were tilted, they would push the surface electrons sideways, causing them to flow. Since the charges are stationary, the field lines must point straight out.

Exam Tip: Mention that "under static conditions, the tangential component of the electric field must be zero."

Question 130. Draw the electric field lines of an isolated point charge Q when (i) Q > 0 and (ii) Q < 0. 
Answer: The configurations are:
(i) **For \( Q > 0 \) (Positive Charge):** The electric field lines are radially directed outwards.
(ii) **For \( Q < 0 \) (Negative Charge):** The electric field lines are radially directed inwards.

In simple words: For positive charges, the field lines shoot straight outward in all directions. For negative charges, the lines point straight inward.

Exam Tip: Ensure you draw arrows indicating direction clearly: pointing outwards for positive and inwards for negative.

Question 131. (i) Depict electric field lines due to two positive charges kept at a certain distance apart. 
(ii) Depict electric field lines due to an electric dipole or due to two opposite charges kept at a certain distance apart.
Answer: The field line configurations are:
(i) **Two Positive Charges:** The field lines repel each other, creating a neutral point directly in the middle with no field lines passing through it.
(ii) **Electric Dipole (Opposite Charges):** The field lines start at the positive charge and curve smoothly to terminate on the negative charge.

In simple words: (i) Two like positive charges push each other's field lines away, leaving an empty spot in the middle. (ii) A positive and a negative charge share field lines that loop directly from the positive to the negative side.

Exam Tip: For like charges, clearly show a "neutral point" (labeled as N) in the center where the net field is zero.

Question 132. (i) A point charge is placed in the vicinity of a conducting surface. Trace the field lines between the charge and the conducting surface. 
(ii) Draw the electric field lines due to uniformly charged thin spherical shell when charge on the shell is (a) positive, (b) negative 
Answer: The patterns are:
(i) **Near a conducting surface:** A positive point charge will induce negative charges on the nearby surface of the conductor. The electric field lines will emerge from the positive charge and terminate perpendicularly on the conductor's surface.
(ii) **Uniformly charged thin spherical shell:**
(a) **Positive Charge:** The field lines point radially outwards from the surface of the shell. Inside the shell, the field is zero, so no lines exist inside.
(b) **Negative Charge:** The field lines point radially inwards, terminating on the shell's surface. Again, no lines exist inside the hollow shell.
In simple words: (i) A charge near a metal plate induces opposite charges on the plate, pulling the field lines straight into the metal. (ii) A charged hollow shell has field lines pointing straight out (if positive) or straight in (if negative), with absolutely no field lines inside the shell.

Exam Tip: Always show that the electric field inside a hollow shell is completely zero by leaving the inside of the circle blank.

Question 133. A metallic sphere is placed in a uniform electric field as shown in the figure. Which path is followed by the electric field lines and why ? 
Answer: The electric field lines will follow **Path 4**.
**Reason:**
1. Electric field lines must always meet and exit the surface of a conductor perpendicularly (normally).
2. The electrostatic field inside a conducting metallic sphere is zero, so no field lines can exist inside the conductor.
Path 4 satisfies both of these conditions.
In simple words: Path 4 is correct because electric field lines must hit the metal ball at a perfect 90-degree angle and cannot pass through the inside of the metal.

Exam Tip: State both reasons (perpendicularity to the surface and zero internal field) to guarantee full marks.

Question 134. Define dipole moment. Write its S.I. unit. Is it a scalar or vector quantity ?
Answer: The electric dipole moment (\( \vec{p} \)) of an electric dipole is defined as the product of the magnitude of either charge (\( q \)) and the distance separating them (\( 2a \)): \[ \vec{p} = q (2\vec{a}) \]
* **SI Unit:** Coulomb-meter (\( \text{C}\cdot\text{m} \)).
* **Type of quantity:** It is a **vector** quantity, directed from the negative charge (\( -q \)) to the positive charge (\( +q \)).
In simple words: Dipole moment measures the strength of a charge pair. It is the charge size times their separation distance, pointing from negative to positive.

Exam Tip: Note the unit abbreviation as \( \text{C}\cdot\text{m} \) (Coulomb-meter) and do not write it as \( \text{cm} \) (centimeter).

Question 135. What is the charge of an electric dipole ?
Answer: The net total charge of an electric dipole is **zero**. This is because a dipole consists of two equal and opposite charges (\( +q \) and \( -q \)), whose algebraic sum is: \[ q_{\text{net}} = +q + (-q) = 0 \]
In simple words: An electric dipole has a total charge of zero because it is made of equal positive and negative charges that cancel each other out.

Exam Tip: Differentiate between "net charge is zero" and "net electric field is not zero."

Question 136. An electric dipole is placed in a uniform electric field, what is the net force acting on it ? 
Answer: The net translational force acting on an electric dipole placed in a uniform electric field is **zero**.
**Reason:** The positive charge experiences a force \( \vec{F}_+ = +q\vec{E} \) along the field, and the negative charge experiences an equal and opposite force \( \vec{F}_- = -q\vec{E} \). The vector sum of these forces is: \[ \vec{F}_{\text{net}} = q\vec{E} + (-q\vec{E}) = 0 \]
In simple words: Because the positive end is pulled forward with the same force that the negative end is pulled backward, the forces cancel out, resulting in no net force.

Exam Tip: Note that although the net force is zero, the net torque is generally non-zero, causing the dipole to rotate.

Question 137. An electric dipole of dipole moment p is placed in a uniform electric field E. Write the value of the angle between p and E for which the torque experienced by the dipole is minimum. 
Answer: The torque \( \tau \) experienced by an electric dipole is: \[ \tau = p E \sin\theta \] The torque is minimum (equal to zero) when \( \sin\theta = 0 \). This occurs when the angle \( \theta \) between \( \vec{p} \) and \( \vec{E} \) is **\( 0^\circ \)** (stable equilibrium) or **\( 180^\circ \)** (unstable equilibrium).
In simple words: The twisting force is zero when the dipole is aligned parallel or antiparallel to the electric field lines.

Exam Tip: State both \( 0^\circ \) and \( 180^\circ \) to provide a complete and accurate answer.

Question 138. Depict the orientation of the dipole in (i) stable, (ii) unstable equilibrium in a uniform electric field. 
Answer: The alignments are:
(i) **Stable Equilibrium (\( \theta = 0^\circ \)):** The dipole moment vector \( \vec{p} \) points in the same direction as the electric field \( \vec{E} \).
(ii) **Unstable Equilibrium (\( \theta = 180^\circ \)):** The dipole moment vector \( \vec{p} \) points in the opposite direction to the electric field \( \vec{E} \).

In simple words: In stable equilibrium, the negative charge is on the left and the positive charge is on the right, pointing with the field. In unstable equilibrium, they are flipped.

Exam Tip: Draw the electric field lines as parallel arrows and clearly show the \( +q \) and \( -q \) positions for both cases.

Question 139. Find the work done in rotating the dipole from stable to unstable equilibrium in a uniform electric field.
Answer: The work done \( W \) in rotating an electric dipole in a uniform electric field from an angle \( \theta_1 \) to \( \theta_2 \) is: \[ W = p E (\cos\theta_1 - \cos\theta_2) \]
For stable equilibrium, the angle is \( \theta_1 = 0^\circ \).
For unstable equilibrium, the angle is \( \theta_2 = 180^\circ \).
Substituting these values: \[ W = p E (\cos(0^\circ) - \cos(180^\circ)) \] \[ W = p E (1 - (-1)) = 2 p E \] The total work done is \( 2 p E \).
In simple words: Twisting the dipole from pointing with the field to pointing against it requires work equal to twice the product of the dipole moment and field strength.

Exam Tip: State the general work formula clearly before plugging in the angles \( 0^\circ \) and \( 180^\circ \).

Question 140. Find the work done in rotating the dipole from unstable to stable equilibrium in a uniform electric field. 
Answer: The work done \( W \) in rotating a dipole from an initial angle \( \theta_1 \) to a final angle \( \theta_2 \) is: \[ W = p E (\cos\theta_1 - \cos\theta_2) \]
Here, rotating from unstable (\( \theta_1 = 180^\circ \)) to stable (\( \theta_2 = 0^\circ \)) equilibrium gives: \[ W = p E (\cos(180^\circ) - \cos(0^\circ)) \] \[ W = p E (-1 - 1) = -2 p E \] The work done by the external agent is \( -2 p E \) (which means the field itself does positive work of \( 2 p E \)).
In simple words: Rotating the dipole to align back with the field releases energy, meaning the work done by us is negative, \( -2 p E \).

Exam Tip: Be careful with the signs; work done *on* the system is negative because the system relaxes to a lower potential energy state.

Question 141. Define electric flux. Write its S.I. unit. 
Answer: Electric flux (\( \Phi_E \)) is defined as the total measure of the number of electric field lines passing normally through a given surface area.
Mathematically, it is the surface integral of the electric field vector \( \vec{E} \) over the area \( \vec{ds} \): \[ \Phi_E = \oint \vec{E} \cdot \vec{ds} \]
* **SI Unit:** Newton-meter squared per Coulomb (\( \text{N}\cdot\text{m}^2/\text{C} \)) or Volt-meter (\( \text{V}\cdot\text{m} \)).
In simple words: Electric flux measures how much electric field is passing through a net surface area, like wind blowing through a window.

Exam Tip: Write the integral formula \( \Phi_E = \oint \vec{E} \cdot \vec{ds} \) to get full marks on definition questions.

Question 142. State Gauss’s law in electrostatics. 
Answer: Gauss's Law states that the total net electric flux (\( \Phi_E \)) passing through any closed imaginary surface (called a Gaussian surface) is equal to \( \frac{1}{\varepsilon_0} \) times the net electric charge (\( q \) ) enclosed within that surface: \[ \Phi_E = \oint \vec{E} \cdot \vec{ds} = \frac{q}{\varepsilon_0} \] where \( \varepsilon_0 \) is the permittivity of free space.
In simple words: Gauss's law says if you enclose some charges in a closed bubble, the total electric field lines coming out of the bubble only depends on the total charge trapped inside, divided by a constant.

Exam Tip: Ensure you specify that the surface must be a "closed surface" for Gauss's law to apply.

Question 143. A charge q is enclosed by a spherical surface R. If the radius is doubled/ reduced to half, how would the electric flux through the surface change ? 
Answer: There will be **no change** in the total electric flux passing through the surface.
**Reason:** According to Gauss's Law, the total electric flux \( \Phi_E = \frac{q}{\varepsilon_0} \) depends solely on the net charge enclosed by the closed surface and is completely independent of the shape, size, or radius of the enclosing boundary.
In simple words: The total flux doesn't change when you scale the sphere because the same amount of charge is still trapped inside, so the same number of field lines must pass through.

Exam Tip: State the mathematical formula \( \Phi_E = q/\varepsilon_0 \) to justify why the radius does not affect the flux.

Question 144. A charge q is placed at the centre of a cube, what is the electric flux passing through one of its faces ? 
Answer: According to Gauss's Law, the total electric flux passing through all six faces of the closed cube is: \[ \Phi_{\text{total}} = \frac{q}{\varepsilon_0} \] Since a cube has six identical symmetric faces and the charge is placed exactly at the center, the flux is distributed equally. Thus, the flux \( \Phi \) through any single face is: \[ \Phi = \frac{1}{6} \Phi_{\text{total}} = \frac{q}{6 \varepsilon_0} \]
In simple words: The total flux coming out of the entire cube is \( q/\varepsilon_0 \). Since a cube has six identical sides, each side gets exactly one-sixth of the total flux.

Exam Tip: Mention "due to symmetry, the flux divides equally among the six faces" to show complete logical reasoning.

Question 145. Consider two hollow concentric spheres, S1 & S2, enclosing charges 2Q & 4Q respectively as shown.
(i) Find out the ratio of the electric flux through them.
(ii) how will the electric flux through the sphere S1 change, if a medium of dielectric constant \varepsilon_r is introduced in the space inside S1 in place of air ? Deduce the necessary expression. 
Answer: The results are:
**(i) Ratio of the Electric Flux:**
The total charge enclosed by the inner sphere \( S_1 \) is: \[ q_1 = 2Q \] The electric flux \( \Phi_1 \) through \( S_1 \) is: \[ \Phi_1 = \frac{2Q}{\varepsilon_0} \] The total charge enclosed by the outer sphere \( S_2 \) is the sum of both charges: \[ q_2 = 2Q + 4Q = 6Q \] The electric flux \( \Phi_2 \) through \( S_2 \) is: \[ \Phi_2 = \frac{6Q}{\varepsilon_0} \] Taking the ratio: \[ \frac{\Phi_1}{\Phi_2} = \frac{\left(\frac{2Q}{\varepsilon_0}\right)}{\left(\frac{6Q}{\varepsilon_0}\right)} = \frac{2}{6} = \frac{1}{3} \] The ratio of the fluxes is \( 1 : 3 \).

**(ii) Effect of introducing a dielectric medium inside \( S_1 \):**
When a medium of dielectric constant \( \varepsilon_r \) replaces air inside the sphere \( S_1 \), the permittivity of the space changes from \( \varepsilon_0 \) to \( \varepsilon = \varepsilon_r \varepsilon_0 \). The new electric flux \( \Phi_1' \) becomes: \[ \Phi_1' = \frac{2Q}{\varepsilon} = \frac{2Q}{\varepsilon_r \varepsilon_0} = \frac{\Phi_1}{\varepsilon_r} \] Thus, the electric flux through the sphere \( S_1 \) decreases by a factor of \( \varepsilon_r \).

In simple words: (i) Sphere 1 encloses 2Q of charge, while Sphere 2 encloses a total of 6Q. Their flux ratio is 2 to 6, which simplifies to 1/3. (ii) Filling the inner space with a dielectric reduces its flux by a factor of the dielectric constant.

Exam Tip: Remember that for the outer sphere \( S_2 \), you must sum up all the charges inside it (\( 2Q + 4Q \)) to calculate the correct flux.

Question 146. (i) Define electric potential energy of a system of charges.
(ii) Write an expression of electric potential energy of a system of two charges. 
Answer: The definitions and relations are:
(i) **Electric Potential Energy:** The electric potential energy of a system of point charges is defined as the total amount of work done by an external agent in assembling the charges at their respective locations from an initial state of infinite separation, without causing acceleration.
(ii) **Expression for Two Charges:** For two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) in a vacuum, the potential energy \( U \) is: \[ U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r} \]
In simple words: (i) It is the total work needed to bring all the charges from incredibly far away and place them next to each other. (ii) For two charges, it is the product of the charges divided by their distance, times a constant.

Exam Tip: Note that potential energy is a scalar quantity and can be positive or negative depending on the signs of \( q_1 \) and \( q_2 \).

Question 147. The figure shows field lines of a positive point charge. What will be the sign of the potential energy deference of a small negative charge between the points Q and P. Justify your answer. 
Answer: Let \( q = -q_0 \) be a small negative test charge, and \( +Q \) be the source charge at the center. The electrostatic potential energy \( U \) at any distance \( r \) is: \[ U = \frac{1}{4 \pi \varepsilon_0} \frac{Q (-q_0)}{r} = -\frac{1}{4 \pi \varepsilon_0} \frac{Q q_0}{r} \] Since point P is closer to the positive source charge than point Q (\( r_P < r_Q \)), the potential energy at P is more negative (lower) than at Q: \[ U_P < U_Q \] Therefore, the potential energy difference \( U_Q - U_P \) is **positive** (\( U_Q - U_P > 0 \)).

In simple words: A negative charge feels a strong, stable attraction close to a positive charge, making its potential energy at P very low (highly negative). Since energy at Q is higher (less negative), the difference \( U_Q - U_P \) is positive.

Exam Tip: Clearly define that "more negative" means a lower algebraic value when explaining potential energy differences.

Question 148. Figure shows the field lines of a negative point charge. Give the sign of the potential energy deference of a small negative charge between the points A and B. 
Answer: Let the source charge be \( -Q \) and the test charge be \( -q_0 \). The electrostatic potential energy \( U \) is positive since both charges are negative: \[ U = \frac{1}{4\pi\varepsilon_0} \frac{(-Q)(-q_0)}{r} = \frac{1}{4\pi\varepsilon_0} \frac{Q q_0}{r} \] Since point A is closer to the negative source charge than point B (\( r_A < r_B \)), the potential energy at A is higher than at B due to the strong repulsion: \[ U_A > U_B \] Therefore, the potential energy difference \( U_A - U_B \) is **positive** (\( U_A - U_B > 0 \)).

In simple words: Pushing two negative charges close together takes work against repulsion, making the energy at A higher. So, subtracting the energy at B from A gives a positive number.

Exam Tip: Note that for like charges, potential energy is positive and increases as distance decreases (\( U \propto \frac{1}{r} \)).

Question 140. The figure shows field lines of a positive point charge. Is the work done by the field in moving a small positive charge from Q to P is positive or negative ? Justify your answer.
Answer: The work done by the electric field is **negative**.
**Justification:**
As a positive test charge \( q_0 \) is moved from point Q to point P closer to the positive source charge \( +Q \), it experiences a repulsive electrostatic force directed away from the source charge. The displacement \( \vec{dr} \) is directed inwards (from Q to P), which is opposite to the electrostatic force \( \vec{F}_e \) pointing outwards. Since the angle between the force and displacement vectors is \( 180^\circ \): \[ W = F_e \cdot dr \cos(180^\circ) < 0 \] Therefore, the work done by the electric field is negative (while the work done by the external agent is positive).
In simple words: The positive source charge repels our test charge. Moving it closer means pushing against the field's natural force, so the electric field does negative work.

Exam Tip: Clearly distinguish between "work done by the electric field" (which is negative here) and "work done by an external agent" (which is positive).

Question 150. The field lines of a negative point charge are as shown in the figure. Does the kinetic energy of a small negative charge increase or decrease in going from B to A ? 
Answer: The kinetic energy of the negative charge **decreases**.
**Reason:**
As the negative charge moves from point B to point A (closer to the negative source charge), it experiences an increasing repulsive electrostatic force. This repulsive force acts in the direction opposite to its motion, decelerating the charge and reducing its velocity. Consequently, its kinetic energy decreases.
In simple words: Since both charges are negative, they repel each other. Moving the negative charge closer to the source charge is like rolling a ball uphill; it slows down, losing kinetic energy.

Exam Tip: Connect this behavior to energy conservation: as potential energy increases due to repulsion, kinetic energy must decrease.

Question 151. (i) Define an equipotential surface ?
(ii) Write any two properties of an equipotential surface. 
Answer: The definitions and properties are:
(i) **Equipotential Surface:** Any surface drawn in an electric field over which the electric potential remains identical at every single point is called an equipotential surface.
(ii) **Properties:**
1. The net work done in moving an electric charge between any two points on an equipotential surface is zero.
2. Electric field lines are always perpendicular (normal) to the equipotential surface at every point.
3. No two equipotential surfaces can intersect each other.
In simple words: (i) An equipotential surface is a boundary where the voltage is completely identical everywhere. (ii) No work is needed to slide charges along this surface, and electric field lines always cross it at a clean 90-degree angle.

Exam Tip: State why two equipotential surfaces can't intersect (it would mean two different potential values at the same point).

Question 152. “For any charge configuration, equipotential surface through a point is normal to the electric field.” Justify this statement. 
Answer: By definition, the potential difference \( dV \) between any two adjacent points separated by a displacement \( \vec{dr} \) on an equipotential surface is zero (\( dV = 0 \)). The relationship between the electric field \( \vec{E} \) and potential difference \( dV \) is: \[ dV = -\vec{E} \cdot \vec{dr} \] Substituting \( dV = 0 \) gives: \[ \vec{E} \cdot \vec{dr} = 0 \implies E \, dr \cos\theta = 0 \] Since \( E \neq 0 \) and \( dr \neq 0 \) on the surface: \[ \cos\theta = 0 \implies \theta = 90^\circ \] This mathematically proves that the electric field \( \vec{E} \) must be perpendicular (normal) to the displacement vector \( \vec{dr} \) along the surface at every point.
In simple words: Since the voltage is the same everywhere on the surface, moving a charge sideways does zero work. This is only possible if the electric force pushes straight out at 90 degrees, perpendicular to our movement.

Exam Tip: Use the dot product proof \( \vec{E} \cdot \vec{dr} = 0 \) to secure a perfect score on this standard justification question.

Question 153. No work done in moving a charge from one point to another on an equipotential surface. Why ?
Answer: The work done \( W \) in moving a charge \( q_0 \) between any two points A and B is: \[ W = q_0 (V_B - V_A) \] By definition, all points on an equipotential surface have the same electric potential, so: \[ V_A = V_B \implies V_B - V_A = 0 \] Substituting this back into the work equation: \[ W = q_0 (0) = 0 \] Therefore, no net work is done in moving a charge along an equipotential surface.
In simple words: Because there is no voltage difference between any two points on the surface, there is no change in potential energy, so the work done is exactly zero.

Exam Tip: Start by writing down the fundamental work-potential relationship \( W = q_0 \Delta V \).

Question 154. Can electric field exist tangential to an equipotential surface ? Give reason. 
Answer: No, an electric field cannot exist tangential to an equipotential surface.
**Reason:** If a tangential component of the electric field existed, it would exert a force on any test charge moving along the surface. This would require a non-zero amount of work to be done to move the charge, which directly contradicts the definition of an equipotential surface where work done must always be zero.
In simple words: No. If the field had a sideways component, we would have to push against it to slide charges along the surface, meaning the work wouldn't be zero anymore.

Exam Tip: Emphasize that a tangential field component would violate the "zero work" rule of equipotential surfaces.

Question 155. Why do the equipotential surfaces due to uniform electric field not intersect each other ?
Answer: If two equipotential surfaces were to intersect, the point of intersection would have two different values of electric potential at the same time. This is physically impossible because a single point in space can only have one unique potential value.
In simple words: If they crossed, a single point would have two different voltages at once, which makes no physical sense.

Exam Tip: Focus on the argument that "potential must be unique at every coordinate point."

Question 156. Why the equipotential surfaces about a single charge are not equidistant ?
OR
Why does the separation between successive equipotential surfaces get wider as the distance from the charges increases ? 
Answer: The relation between electric field \( E \) and the separation between equipotential surfaces \( dr \) for a constant potential difference \( dV \) is: \[ E = -\frac{dV}{dr} \implies dr = -\frac{dV}{E} \implies dr \propto \frac{1}{E} \] For a single isolated point charge, the electric field strength decreases as distance increases (\( E \propto \frac{1}{r^2} \)). Since the field \( E \) weakens further away, the spatial separation \( dr \) between successive equipotential surfaces must become larger to maintain the same potential difference \( dV \). This explains why they are not equidistant and get wider further out.
In simple words: As you move further away, the electric field gets weaker. Because the field is weaker, you have to travel a longer distance to experience the same drop in voltage.

Exam Tip: State the inverse proportionality \( dr \propto \frac{1}{E} \) clearly to back up your description.

Question 157. Draw an equipotential surface in a uniform electric field. 
Answer: In a uniform electric field, the equipotential surfaces are a series of parallel planes that are perpendicular to the electric field lines.

In simple words: For a uniform field pointing sideways, the equipotential surfaces are flat vertical sheets standing perpendicular to the field lines.

Exam Tip: Draw parallel lines for the field and draw vertical dashed lines to represent the flat planes.

Question 158. Draw an equipotential surface and corresponding electric field lines for a single point charge (i) +q (q > 0) (ii) -q(q < 0). 
Answer: The configurations are:
(i) **For \( q > 0 \) (Positive Charge):** The equipotential surfaces are concentric spheres centered on the charge, with field lines pointing radially outwards.
(ii) **For \( q < 0 \) (Negative Charge):** The equipotential surfaces are also concentric spheres, but the field lines point radially inwards.

In simple words: For any point charge, the equipotential surfaces are concentric spheres. The field lines go straight out if the charge is positive, and straight in if negative.

Exam Tip: Draw the concentric circles with dashed lines and the electric field lines as solid intersecting lines with arrowheads.

Question 159. (i) Draw the equipotential surfaces for an electric dipole. 
(ii) Draw the equipotential surfaces due to two equal positive point charges placed at a certain distance. 
Answer: The patterns are:
(i) **For an electric dipole:** The surfaces are closely packed between the charges because the electric field is strongest there.
(ii) **For two equal positive charges:** The surfaces are spaced further apart between the charges because the electric fields oppose each other and weaken in the central region.

In simple words: (i) For a dipole, the voltage spheres squeeze together in the space between the charges. (ii) For two positive charges, they push each other's spheres away, leaving them flatter in the middle.

Exam Tip: Show the distortion of the circles clearly to reflect where the electric field is stronger or weaker.

Question 160. A charge ’ is being moved from a point A above a dipole of dipole moment p to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process.  
Answer: The work done in this process is **zero**.
**Reason:**
The entire equatorial plane of an electric dipole behaves as an equipotential surface where the electric potential \( V \) is constant and equal to zero (\( V = 0 \)) at all points. Since points A and B both lie on this equatorial plane: \[ V_A = V_B = 0 \implies V_B - V_A = 0 \] The work done \( W \) is: \[ W = q (V_B - V_A) = q (0) = 0 \]
In simple words: The equatorial plane of a dipole has a voltage of zero everywhere. Moving a charge along a path where the voltage is always zero requires no work.

Exam Tip: Clearly state that "the equatorial plane of a dipole is an equipotential surface" to justify the zero work value.

Question 161. What is the amount of work done in moving a point charge Q around a circular arc of radius ‘r’ at the centre of which another point charge ‘q’ is located ?
Answer: The work done is **zero**.
**Reason:**
The circular path of radius \( r \) around the central charge \( q \) is a circle of constant radius. Since electric potential is defined as \( V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r} \), all points on this circular arc are at the same distance \( r \) from the source charge, making the path an equipotential line. Since the potential difference \( \Delta V = 0 \): \[ W = Q \Delta V = 0 \]
In simple words: Moving a charge in a circle around another charge keeps it at a constant distance, meaning the voltage never changes. Since the voltage difference is zero, the work done is also zero.

Exam Tip: Identify that a circle around a point charge is a circular equipotential line where potential difference is zero.

Question 162. Define the capacitance of a conductor. Write its S .I. unit. 
Answer: The electrical capacitance \( C \) of a conductor is defined as the ratio of the magnitude of electric charge \( Q \) given to it to the resulting rise in its electrical potential \( V \): \[ C = \frac{Q}{V} \]
* **SI Unit:** **Farad (F)** (where \( 1\text{ Farad} = 1\text{ Coulomb/Volt} \)).
In simple words: Capacitance is the measure of how much electrical charge a conductor can store per Volt of electrical potential. It is measured in Farads.

Exam Tip: Write the basic ratio formula \( C = Q/V \) to support your verbal definition.

Question 163. Define the capacitance of a capacitor. On what factors does it depends ?
Answer: The definition and dependencies are:
The capacitance \( C \) of a capacitor is defined as the ratio of the magnitude of charge \( Q \) on either plate to the potential difference \( V \) established between them: \[ C = \frac{Q}{V} \]
**Factors affecting capacitance:**
(i) **Geometrical Configuration:** The shape, surface area \( A \), and the separation distance \( d \) between the plates.
(ii) **Medium:** The nature of the dielectric medium filled in the space separating the plates.
In simple words: Capacitance is the charge stored per Volt. It depends on how big the plates are, how close they are to each other, and what kind of insulating material is stuffed between them.

Exam Tip: Mention both geometry (area, distance) and the medium to provide a complete answer.

Question 164. Define dielectric constant of a medium in terms of capacitance. 
Answer: The dielectric constant (\( K \) or \( \varepsilon_r \)) of a medium is defined as the ratio of the capacitance (\( C \)) of a capacitor when the space between its plates is completely filled with that dielectric medium to its capacitance (\( C_0 \)) when there is a vacuum (or air) between the plates: \[ K = \frac{C}{C_0} \]
In simple words: The dielectric constant is a factor showing how many times a capacitor's storage capacity increases when you fill the gap with a specific insulating material instead of empty space.

Exam Tip: Express this definition mathematically as \( K = C/C_0 \) to get full marks.

Question 165. A metal plate is introduced between the plates of a charged parallel plate capacitor. What is the effect on the capacitance of the capacitor ? 
Answer: The capacitance of the capacitor **increases**.
**Reason:**
Introducing a conducting metal plate effectively reduces the distance between the plates through which the electric field exists, as the electric field inside the metal plate is zero. Since capacitance is inversely proportional to this effective separation distance (\( C \propto \frac{1}{d-t} \)), reducing this gap increases the capacitance.
In simple words: Inserting a metal plate effectively reduces the empty gap between the capacitor plates, which boosts the capacitor's ability to store charge.

Exam Tip: Note that a metal plate acts as a dielectric with an infinite dielectric constant (\( K = \infty \)), which decreases the potential difference and increases capacitance.

Question 166. (i) Define the term polarization of a dielectric.
(ii) Write a relation for polarization P of a dielectric material in the presence of an external electric field E.
Answer: The relations and definitions are:
(i) **Polarization of a Dielectric:** It is defined as the process of inducing a net electric dipole moment per unit volume of a dielectric material when it is placed inside an external electric field.
(ii) **Relationship:** \[ \vec{P} = \chi_e \varepsilon_0 \vec{E} \] where \( \chi_e \) is the dimensionless electric susceptibility of the dielectric medium and \( \varepsilon_0 \) is the permittivity of free space.
In simple words: (i) Polarization is what happens when an electric field stretches the molecules of an insulator, creating tiny dipoles throughout the material. (ii) The polarization vector is directly proportional to the applied electric field.

Exam Tip: Define \( \chi_e \) as the electric susceptibility of the medium to show complete conceptual knowledge.

Question 167. How is the electric field due to a charged parallel plate capacitor affected when a dielectric slab is inserted between the plates fully occupying the intervening region ? 
Answer: The electric field inside the capacitor **decreases**.
**Reason:**
Inserting the dielectric slab causes polarization, which induces positive and negative bound charges on the opposite faces of the slab. These charges create an internal induced electric field \( \vec{E}_{\text{in}} \) that opposes the external field \( \vec{E}_0 \). The net electric field \( \vec{E} \) is reduced to: \[ E = E_0 - E_{\text{in}} = \frac{E_0}{K} \] where \( K \) is the dielectric constant of the slab.
In simple words: The material polarizes and creates its own small electric field pointing backward, which fights the main field and weakens it by a factor of the dielectric constant.

Exam Tip: Show the equation \( E = E_0/K \) to explain how the field is scaled down.

Question 168. The graph shows the variation of voltage V across the plates of two capacitors A and B versus increase of charge Q stored on them. Which of the capacitors has higher capacitance ? Give reason for your answer. 
Answer: **Capacitor B** has the higher capacitance.
**Reason:**
The capacitance is defined as: \[ C = \frac{Q}{V} \] If we consider a constant voltage line \( V = \text{constant} \), then the capacitance is directly proportional to the charge stored: \[ C \propto Q \] From the graph, at any fixed value of voltage \( V \), the charge stored on capacitor B is greater than that on capacitor A (\( Q_B > Q_A \)). Therefore: \[ C_B > C_A \]

In simple words: For any set voltage, capacitor B holds more electrical charge than A. Since capacitance is the charge stored per Volt, B is the superior capacitor.

Exam Tip: State how the slope of the \( V \text{ vs } Q \) line relates to \( 1/C \) to justify your answer graphically.

Question 169. A parallel plate capacitor of plate area A and separation d is filled with dielectrics of dielectric constants K1 and K2 shown in the figure. Find the net capacitance of the capacitor.
Answer: As shown in the figure, the two dielectric slabs split the gap into two halves along the thickness, each of thickness \( \frac{d}{2} \) and area \( A \). This setup is equivalent to two capacitors, \( C_1 \) and \( C_2 \), connected in series.
The individual capacitances are: \[ C_1 = \frac{K_1 \varepsilon_0 A}{d/2} = \frac{2 K_1 \varepsilon_0 A}{d} \] \[ C_2 = \frac{K_2 \varepsilon_0 A}{d/2} = \frac{2 K_2 \varepsilon_0 A}{d} \] Since they are connected in series, the net equivalent capacitance \( C \) is: \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \] \[ \frac{1}{C} = \frac{d}{2 K_1 \varepsilon_0 A} + \frac{d}{2 K_2 \varepsilon_0 A} = \frac{d}{2 \varepsilon_0 A} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) \] \[ \frac{1}{C} = \frac{d}{2 \varepsilon_0 A} \left( \frac{K_1 + K_2}{K_1 K_2} \right) \] \[ C = \left( \frac{2 K_1 K_2}{K_1 + K_2} \right) \frac{\varepsilon_0 A}{d} = \left( \frac{2 K_1 K_2}{K_1 + K_2} \right) C_0 \] where \( C_0 = \frac{\varepsilon_0 A}{d} \) is the capacitance in air.

In simple words: Slicing the gap horizontally creates two capacitors connected in series. Calculating their combined series resistance gives a final capacitance formula using the harmonic mean of the two dielectric constants.

Exam Tip: Identify that splitting along the thickness \( d \) yields a series combination, whereas splitting along the area \( A \) yields a parallel combination.

Question 170. Two dielectric slabs of dielectric constants K1 and K2 are filled in between the two plates, each of area A, of the parallel plate capacitor as shown. Find net capacitance of the capacitor. 
Answer: Here, the two dielectric slabs are placed side-by-side, splitting the total area into two equal halves \( \frac{A}{2} \) while keeping the plate separation \( d \) identical. This setup is equivalent to two capacitors, \( C_1 \) and \( C_2 \), connected in parallel.
The individual capacitances are: \[ C_1 = \frac{K_1 \varepsilon_0 (A/2)}{d} = \frac{K_1 \varepsilon_0 A}{2d} \] \[ C_2 = \frac{K_2 \varepsilon_0 (A/2)}{d} = \frac{K_2 \varepsilon_0 A}{2d} \] Since they are connected in parallel, the net equivalent capacitance \( C \) is: \[ C = C_1 + C_2 \] \[ C = \frac{K_1 \varepsilon_0 A}{2d} + \frac{K_2 \varepsilon_0 A}{2d} = \frac{\varepsilon_0 A}{2d} (K_1 + K_2) \] \[ C = \left( \frac{K_1 + K_2}{2} \right) \frac{\varepsilon_0 A}{d} = \left( \frac{K_1 + K_2}{2} \right) C_0 \] where \( C_0 = \frac{\varepsilon_0 A}{d} \) is the capacitance in air.

In simple words: Slicing the gap vertically creates two capacitors in parallel. The final capacitance is simply the average of the two dielectric factors multiplied by the original air capacitance.

Exam Tip: Remember that parallel capacitors add up directly, which makes the final multiplier equal to the arithmetic mean of \( K_1 \) and \( K_2 \).

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