CBSE Class 12 Physics Electrostat Assignment Set 01

Electrostatics HOTS
 

Question. Where does the electrostatic energy of a parallel plate capacitor reside?
Answer: The energy is stored within the dielectric medium located between the two plates.
In simple words: The electrical energy of a capacitor is kept in the insulating material between its plates.

Exam Tip: Remember to mention "dielectric medium" specifically as the region where the electrostatic field energy is stored.

Question. Do electrons tend to move to regions of lower or higher potential, and why?
Answer: Electrons travel towards the region of higher potential because they carry a negative electric charge.
In simple words: Since electrons have a negative charge, they are naturally attracted to higher potential areas.

Exam Tip: Always state that negative charges move from lower potential to higher potential, whereas positive charges do the opposite.

Question. What is the total net charge on a charged parallel plate capacitor?
Answer: The total net charge on a parallel plate capacitor is zero.
In simple words: One plate is positive and the other is negative by the exact same amount, so they cancel out to give zero total charge.

Exam Tip: Differentiate clearly between the charge on a single plate (which is \( Q \)) and the net charge of the entire capacitor system (which is \( 0 \)).

Question. A Gaussian surface encloses an electric dipole within it. What is the total electric flux across the sphere?
Answer: The net electric flux is zero because the net charge enclosed by the surface is zero.
In simple words: A dipole has equal positive and negative charges, which add up to zero. Thus, no net electric flux passes out of the surface.

Exam Tip: Apply Gauss's law \( \phi = \frac{q_{\text{en}}}{\varepsilon_0} \) and show that the enclosed charge \( q_{\text{en}} = +q - q = 0 \).

Question. Find the dimensional formula of the electrostatic energy density \( \frac{1}{2}\varepsilon_0 E^2 \).
Answer: The dimensions of electrostatic energy density (energy per unit volume) are given by \( \text{ML}^{-1}\text{T}^{-2} \).
In simple words: Energy density means energy divided by volume, which has the dimensional formula of pressure.

Exam Tip: Derive the dimensional formula using the units of energy (Joules) and volume (cubic meters) to avoid memorization errors.

Question. In a certain \( 1\text{ m}^3 \) region of space, the electric potential is found to be \( V\text{ Volts} \) throughout. What is the electric field in this region?
Answer: The electric field in this region is zero.
In simple words: Since the potential is constant and does not change from point to point, the electric field is zero.

Exam Tip: Use the relation \( E = -\frac{dV}{dr} \). Since \( V \) is constant, its derivative is zero, which proves \( E = 0 \).

Question. If Coulomb's law involved a \( \frac{1}{r^3} \) dependence instead of a \( \frac{1}{r^2} \) dependence, would Gauss's law still be true?
Answer: No, Gauss's law would no longer remain valid.
In simple words: Gauss's law depends on the inverse-square nature of Coulomb's law, so any other power of distance would make it invalid.

Exam Tip: State that the proof of Gauss's law relies on the inverse-square law, which cancels the \( r^2 \) term from the area of a sphere.

Question. Why do electrostatic field lines not form continuous closed loops?
Answer: Field lines act as indicators of the electric field, which originates from positive charges and terminates at negative charges, extending to an infinite distance for isolated charges rather than looping back.
In simple words: Electric field lines have starting and ending points on charges, so they cannot loop back onto themselves like magnetic lines do.

Exam Tip: Explain this by stating that the electrostatic field is conservative in nature, which mathematically prevents the formation of closed loops.

Question. The given graph shows the variation of charge versus potential difference \( V \) for two capacitors \( C_1 \) and \( C_2 \). The two capacitors have the same plate separation, but the plate area of \( C_2 \) is doubled compared to \( C_1 \). Which of the lines in the graph corresponds to \( C_1 \) and \( C_2 \), and why?
Answer: The capacitance \( C \) is directly proportional to the plate area \( A \). Therefore, we have \( C_2 = 2C_1 \). Since \( C = \frac{Q}{V} \), the slope of the \( Q-V \) graph represents the capacitance. Consequently, the line with the steeper slope \( P \) represents \( C_2 \), and the line \( Q \) represents \( C_1 \).
In simple words: A larger plate area means a capacitor can hold more charge. On a graph of charge against voltage, the steeper line shows the capacitor with the larger capacity.

Exam Tip: Check which quantity is on the y-axis and which is on the x-axis before concluding whether the slope represents \( C \) or \( \frac{1}{C} \).

Question. Three charges, each equal to \( +2\text{ C} \), are placed at the corners of an equilateral triangle. If the force between any two charges is \( F \), what will be the net force on any one charge?
Answer: Each charge experiences two forces, each of magnitude \( F \), oriented at an angle of \( 60^\circ \) relative to each other. Using the vector addition formula, their resultant force is: \[ F_{\text{net}} = \sqrt{F^2 + F^2 + 2F^2 \cos(60^\circ)} = \sqrt{3}F \]
In simple words: Since the charges are at the corners of an equilateral triangle, the angle between the two pushing forces is 60 degrees. Adding these forces together as vectors gives a total force that is \( \sqrt{3} \) times a single force.

Exam Tip: Clearly show the vector addition steps and use \( \cos(60^\circ) = 0.5 \) to calculate the correct factor of \( \sqrt{3} \).

Question. A point charge \( q \) is placed at \( O \). Is \( V_P - V_Q \) positive or negative when (i) \( q > 0 \) and (ii) \( q < 0 \)? Justify your answer.
Answer:
(i) According to the definition of potential difference, when \( q > 0 \), we have \( V_P > V_Q \). Thus, \( V_P - V_Q \) is positive.
(ii) When \( q < 0 \), the potential values are negative, giving \( V_Q > V_P \). Thus, \( V_P - V_Q \) is negative.
In simple words: For a positive charge, potential drops as you move further away, so the closer point is at a higher potential. For a negative charge, the closer point is more negative, which means it is at a lower potential.

Exam Tip: Write the potential formula \( V = \frac{kq}{r} \) and show how the sign of \( q \) reverses the inequality when comparing \( r_P \) and \( r_Q \).

Question. An electric dipole of dipole moment \( 20 \times 10^{-6}\text{ C}\cdot\text{m} \) is enclosed by a closed surface. What is the net flux coming out of the surface?
Answer: The net flux is zero because the total enclosed charge \( q_{\text{en}} \) of the dipole is zero.
In simple words: The dipole has equal positive and negative charges inside the surface, so they cancel out and no net flux comes out.

Exam Tip: State Gauss's law and explicitly show that the sum of the charges of a dipole is zero.

Question. Why does the electric field inside a dielectric decrease when it is placed in an external electric field?
Answer: Due to electric polarization, an internal opposing electric field is created within the dielectric, which partially cancels the external field.
In simple words: The external field shifts the charges inside the material, creating a small backward field that makes the overall field inside weaker.

Exam Tip: Use the relation \( E_{\text{net}} = E_0 - E_p \), where \( E_0 \) is the applied field and \( E_p \) is the polarization field.

Question. Write the magnitude and direction of the electric field intensity due to an electric dipole of length \( 2a \) at the midpoint of the line joining the two charges.
Answer: The electric field at the midpoint is \( \frac{2kq}{a^2} \). It points towards the negative charge, which is in the direction opposite to the electric dipole moment vector.
In simple words: At the center, the positive charge pushes and the negative charge pulls in the same direction, so their fields add up to point towards the negative charge.

Exam Tip: Be careful with the distance; since the dipole length is \( 2a \), the distance from either charge to the midpoint is \( a \).

Question. A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume. What is the electric field inside the empty space (cavity)?
Answer: The electric field at any point inside the cavity is uniform and non-zero.
In simple words: Removing a sphere of charge leaves an empty pocket where the electric field has the same strength and direction everywhere.

Exam Tip: Explain this using the principle of superposition, combining the fields of a fully charged sphere and a oppositely charged smaller sphere.

Question. A charged particle is free to move in an electric field. Will it always move along an electric line of force?
Answer: No, it will not. If the initial velocity of the charged particle is directed at an angle to the line of force, the particle will not travel along the field line.
In simple words: If a particle is already moving in another direction when it enters the field, it will follow a curved path instead of sticking directly to the field lines.

Exam Tip: Clarify that field lines indicate the direction of force (acceleration), not necessarily the velocity or the path of motion.

Question. If \( V = \frac{q}{4\pi\varepsilon_0 r} \) is the potential at a distance \( r \) due to a point charge \( q \), then determine the electric field using the relation between field and potential.
Answer: The electric field is derived as: \[ E = -\frac{dV}{dr} = -\frac{d}{dr}\left( \frac{q}{4\pi\varepsilon_0 r} \right) = \frac{q}{4\pi\varepsilon_0 r^2} \]
In simple words: Taking the rate of change of voltage with distance gives the electric field strength, which matches Coulomb's law.

Exam Tip: Show the differentiation steps clearly, noting that the derivative of \( \frac{1}{r} \) is \( -\frac{1}{r^2} \).

Question. Is there any point in space where the electric field is non-zero but the electric potential is zero? Give an example.
Answer: Yes, this occurs at any point on the equatorial plane of an electric dipole.
In simple words: In the exact middle of a dipole, the positive and negative voltages cancel out to zero, but the electric field is still strong and active.

Exam Tip: Mention the equatorial point of an electric dipole as a classic example of this condition.

Question. Devise an arrangement of three point charges separated by finite distances that has zero total electrostatic potential energy.
Answer: An arrangement of three collinear charges where a positive charge \( q \) lies at the center, with two negative charges of the same magnitude positioned at equal distances \( a \) on either side, has zero total energy: \[ U = \frac{kq^2}{a} - \frac{kq^2}{2a} - \frac{kq^2}{2a} = 0 \]
In simple words: Putting one positive charge in the middle of two negative charges can balance out the pushing and pulling energies to exactly zero.

Exam Tip: Set up the potential energy equation for all three pairs of charges and show that the sum adds up to zero.

Question. Each of the uncharged capacitors in the circuit has a capacitance of \( 25\ \mu\text{F} \). What charge shall flow through the meter \( M \) when the switch \( S \) is closed to a source of \( 4200\text{ V} \)?
Answer: Since the three capacitors are connected in parallel, their equivalent capacitance is: \[ C_{\text{net}} = C + C + C = 3C = 3 \times 25\ \mu\text{F} = 75\ \mu\text{F} \] The total charge that flows through the circuit is: \[ Q = C_{\text{net}} \times V = 75\ \mu\text{F} \times 4200\text{ V} = 315\text{ mC} \]
In simple words: In parallel, the capacities add up to \( 75\ \mu\text{F} \). Multiplying this total capacity by the voltage of \( 4200\text{ V} \) gives a total charge of \( 315\text{ mC} \).

Exam Tip: State the parallel combination formula clearly before performing the multiplication.

Question. A charge of \( 2\text{ C} \) is placed at the center of a cube of volume \( 8\text{ cm}^3 \). What is the electric flux passing through one face of the cube?
Answer: The total flux through the cube is \( \phi = \frac{q}{\varepsilon_0} = \frac{2}{\varepsilon_0} \). Since a cube has six identical faces, the flux through any single face is: \[ \phi_{\text{face}} = \frac{\phi}{6} = \frac{1}{3\varepsilon_0} \]
In simple words: The total flux from the charge divides equally among the six faces of the cube, so each face gets one-sixth of the total flux.

Exam Tip: Note that the volume of the cube is extra information and does not affect the calculation of the flux.

Question. A charged particle \( q \) is shot towards a fixed charge \( Q \) with speed \( v \), approaching up to a minimum distance \( r \) before returning. If the initial speed was \( 2v \), what would be the new distance of closest approach?
Answer: At the closest distance, kinetic energy is fully converted into potential energy: \[ \frac{1}{2}mv^2 = \frac{kQq}{r} \implies v^2 \propto \frac{1}{r} \implies r \propto \frac{1}{v^2} \] If the speed is doubled to \( 2v \), the new distance \( r' \) becomes: \[ r' = \frac{r}{4} \]
In simple words: If you throw the particle twice as fast, it has four times the energy, allowing it to squeeze four times closer to the fixed charge.

Exam Tip: Show the inverse-square relationship between the closest approach distance and the velocity to justify the final answer.

Question. Two capacitors of capacitances \( 6\ \mu\text{F} \) and \( 12\ \mu\text{F} \) are connected in series. If the potential difference across the \( 6\ \mu\text{F} \) capacitor is \( 2\text{ V} \), find the potential difference across the \( 12\ \mu\text{F} \) capacitor and the total voltage of the battery.
Answer: The charge on the first capacitor is: \[ Q = C_1 V_1 = 6\ \mu\text{F} \times 2\text{ V} = 12\ \mu\text{C} \] Since they are connected in series, the same charge flows through both capacitors. The potential difference across the second capacitor is: \[ V_2 = \frac{Q}{C_2} = \frac{12\ \mu\text{C}}{12\ \mu\text{F}} = 1\text{ V} \] The total battery voltage is: \[ V = V_1 + V_2 = 2\text{ V} + 1\text{ V} = 3\text{ V} \]
In simple words: In a series setup, both capacitors store the same amount of charge. Using this, the second capacitor has a voltage of \( 1\text{ V} \), making the total battery voltage \( 3\text{ V} \).

Exam Tip: State the charge conservation rule for series capacitors as the key step of your explanation.

Question. A parallel plate capacitor has a capacitance of \( 10\text{ pF} \) with air between the plates. What will be its capacitance if the plate separation is halved and a medium of dielectric constant \( 4 \) is introduced between them?
Answer: The initial capacitance is \( C_0 = \frac{\varepsilon_0 A}{d} = 10\text{ pF} \). When the separation is halved and a dielectric of constant \( K = 4 \) is added: \[ C_{\text{final}} = \frac{K\varepsilon_0 A}{d/2} = 2K C_0 = 2 \times 4 \times 10\text{ pF} = 80\text{ pF} \]
In simple words: Halving the distance doubles the capacity, and adding the dielectric multiplies it by 4. Thus, the new capacity is 8 times the original value, giving \( 80\text{ pF} \).

Exam Tip: Write the general formula for a capacitor with a dielectric, showing how both changes combine to scale the capacity.

Question. Five identical capacitors are connected in series. If their net equivalent capacitance is \( 5\ \mu\text{F} \), calculate the capacitance of each individual capacitor.
Answer: For five identical capacitors of value \( C \) in series, we have: \[ \frac{1}{C'} = \frac{5}{C} \implies C' = \frac{C}{5} \] Given \( C' = 5\ \mu\text{F} \): \[ C = 5 \times 5\ \mu\text{F} = 25\ \mu\text{F} \]
In simple words: Connecting five equal capacitors in series reduces the overall capacity to one-fifth of a single capacitor's value. Since the net value is \( 5\ \mu\text{F} \), each must be \( 25\ \mu\text{F} \).

Exam Tip: Be careful to invert the fraction correctly when solving series capacitor problems.

Question. When a capacitor is charged to potential \( V \) by a battery, a charge \( q \) flows. Why is the stored potential energy only \( \frac{1}{2}qV \)? What happens to the rest of the energy?
Answer: The work done by the battery in moving charge \( q \) is \( W = qV \). However, the energy stored in the capacitor is only \( U = \frac{1}{2}qV \). The remaining half of the energy (\( qV - \frac{1}{2}qV = \frac{1}{2}qV \)) is lost as heat in the connecting wires during the charging process.
In simple words: The battery does work to charge the capacitor, but half of this energy is wasted as heat in the wires, leaving only half stored inside the capacitor.

Exam Tip: Clearly state that \( 50\% \) of the energy supplied by the battery is inevitably dissipated as heat, regardless of the resistance of the wires.

Question. What is the angle between the electric dipole moment and the electric field strength at any point on the equatorial line of an electric dipole?
Answer: The angle is \( 180^\circ \) because the electric field on the equatorial line is oriented in the direction opposite to the electric dipole moment.
In simple words: On the side of a dipole, the electric field points backward compared to the dipole's internal direction, making the angle between them a straight line of 180 degrees.

Exam Tip: Use a small sketch to show the directions of the dipole moment vector (negative to positive) and the equatorial electric field vector.

Question. Find the equivalent capacitance of a network containing two parallel branches, where each branch consists of a \( 2C \) capacitor connected in series with a \( C \) capacitor.
Answer: The capacitance of a single branch with \( 2C \) and \( C \) in series is: \[ C_{\text{branch}} = \frac{2C \times C}{2C + C} = \frac{2}{3}C \] Since the two identical branches are in parallel, the total equivalent capacitance is: \[ C_{\text{net}} = C_{\text{branch}} + C_{\text{branch}} = \frac{2}{3}C + \frac{2}{3}C = \frac{4}{3}C \]
In simple words: First calculate the series combination of each branch to get \( \frac{2}{3}C \). Then, add the two branches together because they are in parallel to get \( \frac{4}{3}C \).

Exam Tip: Solve step-by-step: reduce the series branches first, then combine the parallel components.

Question. Eight identical spherical liquid drops, each having radius \( r \), merge to form a single larger drop of radius \( R \). By what factor do the radius and potential change?
Answer: Since volume is conserved: \[ \frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3 \implies R = 2r \] The total charge on the larger drop is \( Q = 8q \). The new potential is: \[ V' = \frac{kQ}{R} = \frac{k(8q)}{2r} = 4\left(\frac{kq}{r}\right) = 4V \]
In simple words: Merging 8 drops doubles the radius, while the combined charge raises the surface potential to four times its original value.

Exam Tip: Keep the volume conservation step explicit as it is essential for finding the relation between the radii.

Question. A uniform electric field of \( 2\text{ kV/m} \) is directed along the x-axis. Find the potential difference between two points separated by \( 4\text{ m} \) along this axis.
Answer: The potential difference is calculated as: \[ dV = -E \cdot dx = -2 \times 10^3\text{ V/m} \times 4\text{ m} = -8 \times 10^3\text{ V} \]
In simple words: Moving along the electric field means potential drops. Over a distance of 4 meters in a field of \( 2000\text{ V/m} \), the potential decreases by \( 8000\text{ Volts} \).

Exam Tip: Do not omit the negative sign, which indicates that electric potential decreases in the direction of the electric field.

Question. Two identical metal plates are given charges \( Q_1 \) and \( Q_2 \) respectively, where \( Q_2 < Q_1 \). If they are brought close to form a parallel plate capacitor of capacitance \( C \), find the potential difference between them.
Answer: The electric field between the plates is due to the difference in charges: \[ E = \frac{Q_1 - Q_2}{2A\varepsilon_0} \] The potential difference between the plates is: \[ V = E \cdot d = \frac{Q_1 - Q_2}{2A\varepsilon_0/d} = \frac{Q_1 - Q_2}{2C} \]
In simple words: The potential difference depends on the difference between the charges of the two plates, divided by twice the capacitance.

Exam Tip: Remember that the charges redistribute, causing a net electric field proportional to \( (Q_1 - Q_2) \).

Question. Three charges \( Q \), \( +q \), and \( +q \) are placed at the vertices of a right-angled isosceles triangle of side \( a \). Find the value of \( Q \) for which the total electrostatic potential energy of the configuration is zero.
Answer: The potential energy of the system is the sum of the energies of all three pairs: \[ U = k\left( \frac{q^2}{a} + \frac{qQ}{a} + \frac{qQ}{\sqrt{2}a} \right) \] Setting \( U = 0 \): \[ q + Q\left( 1 + \frac{1}{\sqrt{2}} \right) = 0 \implies Q = -\frac{2q}{2 + \sqrt{2}} \]
In simple words: By writing down the potential energy of all three interactions and setting the sum to zero, we find that \( Q \) must be \( -\frac{2q}{2 + \sqrt{2}} \).

Exam Tip: Pay close attention to the distance between the two \( q \) charges, which is the hypotenuse \( \sqrt{2}a \).

Question. An infinite number of charges, each of magnitude \( q \), are placed along the x-axis at \( x = 1\text{ m} \), \( x = 2\text{ m} \), \( x = 4\text{ m} \times x = 8\text{ m} \), and so on. Find the electric field at \( x = 0 \) due to these charges.
Answer: The net electric field is the sum of the fields of each individual charge: \[ E = \frac{q}{4\pi\varepsilon_0}\left[ 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots \right] \] This forms an infinite geometric progression with first term \( a = 1 \) and common ratio \( r = \frac{1}{4} \). \[ \text{Sum} = \frac{1}{1 - 1/4} = \frac{4}{3} \] Thus, the electric field is: \[ E = \frac{q}{4\pi\varepsilon_0} \times \frac{4}{3} = \frac{q}{3\pi\varepsilon_0} \]
In simple words: Adding up the electric fields of all the charges forms an infinite series. Summing this series mathematically gives a net electric field of \( \frac{q}{3\pi\varepsilon_0} \).

Exam Tip: State the sum formula of an infinite geometric progression \( S_{\infty} = \frac{a}{1-r} \) clearly to show how the series is evaluated.

Question. A charge \( Q \) is distributed over two concentric hollow spheres of radii \( r \) and \( R \) (\( R > r \)) such that their surface charge densities are equal. Find the potential at the common center.
Answer: Let \( q \) and \( q' \) be the charges on the inner and outer spheres. Since their surface charge densities are equal: \[ \frac{q}{4\pi r^2} = \frac{q'}{4\pi R^2} \implies q = \frac{Q r^2}{R^2 + r^2} \quad \text{and} \quad q' = \frac{Q R^2}{R^2 + r^2} \] The potential at the common center \( O \) is: \[ V_0 = \frac{q}{4\pi\varepsilon_0 r} + \frac{q'}{4\pi\varepsilon_0 R} = \frac{Q(R + r)}{4\pi\varepsilon_0(R^2 + r^2)} \]
In simple words: Since the charge densities are equal, we can find the charge on each sphere. Adding their potential contributions at the center gives the final potential.

Exam Tip: First relate the charges using the equal density condition before substituting them into the potential equation.

Question. An electric dipole is held in a uniform electric field. (i) Show that it does not undergo any translatory motion. (ii) Derive the expression for the torque acting on it and specify its direction.
Answer:
(i) A dipole consists of two equal and opposite charges. In a uniform electric field, they experience equal and opposite electrostatic forces (\( +qE \) and \( -qE \)). Since the net force is zero, there can be no translatory motion.
(ii) The torque is given by the product of the force and the perpendicular distance between the forces: \[ \tau = F \times r_{\perp} = qE \times 2l\sin\theta = pE\sin\theta \implies \vec{\tau} = \vec{p} \times \vec{E} \] The direction of the torque is perpendicular to the plane containing \( \vec{p} \) and \( \vec{E} \), pointing outward from the surface.
In simple words: (i) The forces on the positive and negative ends cancel out, so the dipole does not slide. (ii) However, because the forces act at different ends, they twist the dipole with a torque of \( pE\sin\theta \).

Exam Tip: Clearly distinguish between uniform and non-uniform fields; in a non-uniform field, the net force is not zero.

Question. The electrostatic potential inside a charged ball depends only on the distance from its center as \( V = ar^2 + b \), where \( a \) and \( b \) are constants. Find the space charge distribution \( \rho(r) \) inside the ball.
Answer: The electric field is related to the potential by: \[ E = -\frac{dV}{dr} = -2ar \] Using Gauss's law for a spherical surface of radius \( r \): \[ E \cdot 4\pi r^2 = \frac{q_{\text{en}}}{\varepsilon_0} = \frac{1}{\varepsilon_0}\left( \frac{4\pi}{3}r^3\rho \right) \] Substitute \( E = -2ar \): \[ -2ar \cdot 4\pi r^2 = \frac{4\pi}{3\varepsilon_0}r^3\rho \implies \rho = -6a\varepsilon_0 \]
In simple words: First find the electric field by differentiating the potential. Then, apply Gauss's law to relate this field to the charge density, which gives a constant value of \( -6a\varepsilon_0 \).

Exam Tip: Remember to express Gauss's law in its integral form to easily relate the variable electric field to the enclosed volume charge.

PHYSICS TEST SERIES

Hr ELECTROSTAT 

Q1.Chrge q is placed at the centre of a cube. Find the flux passing through the two opposite faces of the cube.

Q2.Potential on the surface of a thin charge spherical shell is 10V.Find the potential at the centre of the shell.

Q3. Q A Two capacitors C1 and C2 having equal distance B between the plates and area of one is twice of other V identify the corresponding graph.

Q4. Draw the equipotential surfaces for q>0.Are they equidistance ,if not write the reason.

Q5. A capacitor of 4μf is charged by 200V.It is then disconnected from supply and is connected to another uncharged capacitor of 2μf.How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Q6. If there is an arc of radius R makes an angle α at its centre having the linear charge density λ. Find the potential at the centre of arc.

Q7.There is an oil drop of radius R in equilibrium between the two plates of the capacitor which are having surface charge densities +σ and –σ respectively and at distance d apart. Density of oil is ρ. Find an expression for excess number of electrons in oil drop.

Q8. A capacitor is charged and then it is disconnected from the source and then distance between its two plates is increased to twice and then a dielectric of constant K is filled between the plates. Find the change in energy stored in the capacitor.

Q9.Find the frequency of oscillations of a dipole of dipole moment p and having rotational inertia I, in a uniform electric field E.

Please refer to the link below for CBSE Class 12 Physics Electrostat Assignment Set A