A parallel plate capacitor with a dielectric constant $K$ is charged to a potential $V$. If the dielectric is removed while the capacitor remains connected to the battery, what happens to the charge on the capacitor?

The charge on the capacitor remains the same.

The charge on the capacitor increases.

The charge on the capacitor decreases.

The charge on the capacitor remains the same.

The charge on the capacitor becomes zero.

Which of the following statements about equipotential surfaces is true?

Equipotential surfaces are perpendicular to electric field lines.

Equipotential surfaces are always parallel to electric field lines.

Equipotential surfaces are perpendicular to electric field lines.

Equipotential surfaces can intersect each other.

Equipotential surfaces have varying potential values at different points.

A charge is moved from point A to point B along an equipotential surface. What is the work done by the electric field?

Work done depends on the path taken.

A charge-free cavity is present inside a conductor. If an external electric field is applied, what will be the electric field inside the cavity?

The electric field will be zero.

The electric field will be equal to the external field.

The electric field will be zero.

The electric field will be half of the external field.

The electric field will be twice the external field.

Which of the following statements correctly describes the behavior of polar and non-polar dielectrics in an external electric field?

Polar dielectrics have permanent dipoles that align with the external field.

Non-polar dielectrics develop a permanent dipole moment in an external field.

Polar dielectrics have permanent dipoles that align with the external field.

Non-polar dielectrics have permanent dipoles that align with the external field.

Polar dielectrics develop induced dipoles in an external field.

In a system of two point charges $q_1$ and $q_2$ separated by a distance $r_{12}$, what is the expression for the potential energy of the system?

$U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}}$

$U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}^2}$

$U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}^3}$

$U = \frac{1}{4\pi\varepsilon_0} q_1 q_2 r_{12}$

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CBSE

Class 12 Science (PCM)

Physics Part 1

Electrostatic Potential and Capacitance

Study Action Plan (Sorted by Difficulty)

Energy Density of Electric Field

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Total Correct0

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Difficulty Level:

High

Energy Stored in Capacitor

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Total Incorrect2

Difficulty Level:

High

Polar and Non-polar Dielectrics

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Learning Objectives

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Learning Objectives:
- Understand the concept of electrostatic potential and its relation to electric fields.
- Calculate the electric potential due to point charges and charge distributions.
- Analyze the behavior of capacitors in series and parallel configurations.
- Apply the principles of electrostatics to solve problems involving capacitors and electric fields.
- Explore the effects of dielectric materials on capacitance and electric fields.
- Investigate the concept of electrostatic shielding and its applications.

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Revision Notes & Summary

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Electrostatic Potential and Capacitance

Introduction

The notion of potential energy was introduced in previous chapters, where work done against a force results in stored potential energy.
Coulomb force between stationary charges is a conservative force, similar to gravitational force.

Key Concepts

Electrostatic Shielding: A cavity inside a conductor is shielded from outside electrical influences, but charges inside the cavity do not shield the exterior.

Exercises

Electric Potential Zero: Two charges 5 X 10⁻⁸ C and -3 X 10⁻⁸ C are located 16 cm apart. Find points where electric potential is zero.
Hexagon Charge Potential: A regular hexagon with side 10 cm has a charge of 5 µC at each vertex. Calculate the potential at the center.
Equipotential Surface: Two charges 2 µC and -2 µC are placed 6 cm apart. Identify an equipotential surface and the direction of the electric field on this surface.
Spherical Conductor: A spherical conductor of radius 12 cm has a charge of 1.6 X 10⁻⁷C. Calculate the electric field inside, just outside, and at a point 18 cm from the center.
Capacitance Calculation: A parallel plate capacitor with air has a capacitance of 8 pF. Calculate capacitance if the distance is halved and filled with a dielectric constant of 6.
Series Capacitors: Three capacitors of 9 pF each are connected in series. Find total capacitance and potential difference across each capacitor connected to a 120 V supply.
Parallel Capacitors: Three capacitors of 2 pF, 3 pF, and 4 pF are connected in parallel. Find total capacitance and charge on each capacitor connected to a 100 V supply.
Capacitance of Parallel Plate: For a parallel plate capacitor with area 6 X 10⁻³ m² and distance 3 mm, calculate capacitance and charge when connected to a 100 V supply.

Important Formulas

Capacitance: C = Q/V, where Q is charge and V is potential difference.
Capacitance in Series: $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdots$
Capacitance in Parallel: C = C₁ + C₂ + C₃ + ...
Energy Stored in Capacitor: U = $\frac{1}{2} C V^2$
Electric Field in Dielectric: E = E₀/K, where K is the dielectric constant.

Points to Ponder

Electrostatics deals with forces between charges at rest, maintained by unspecified opposing forces.
A capacitor confines electric field lines, resulting in a small potential difference despite strong fields.
Electric field is zero inside a charged spherical shell, while potential is continuous across the surface.
The torque on a dipole in an electric field causes oscillation, aligning with the field if damped.
Potential due to a charge at its own location is infinite.

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Practice Test – MCQs, True/False

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Multiple Choice Questions

0

\frac{Q}{4\pi\varepsilon_0 R}

\frac{Q}{4\pi\varepsilon_0 R} - E \cdot R

E \cdot R

Correct Answer: D

Solution:

The potential inside the shell is constant and equal to the potential at the surface. The potential difference between the center and just outside the shell is due to the external field, which is

E \cdot R

, where

E

is the external electric field.

Chapter Concept:

Potential of Uniformly Charged Spherical Shell

The potential is zero.

The potential is negative.

The potential is positive and given by

V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r}

The potential is undefined.

Correct Answer: C

Solution:

For a positive point charge

Q

at the origin, the potential at a point

P

at a distance

r

is positive and given by

V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r}

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Electrostatic Potential and Capacitance

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True or False

Correct Answer: False

Solution:

The potential at a point due to a system of point charges is the algebraic scalar sum of the potentials due to individual charges, not vector addition.

Chapter Concept :

Potential Due to a System of Charges

Correct Answer: True

Solution:

In a static situation, the free charges in a conductor distribute themselves such that the electric field is zero everywhere inside the conductor.

Chapter Concept :

Electrostatics of Conductors

Correct Answer: True

Solution:

Equipotential surfaces are defined as surfaces where the potential is constant. Since no work is done in moving a charge along an equipotential surface, the electric field must be perpendicular to it.

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Electrostatic Potential and Capacitance

Study Action Plan (Sorted by Difficulty)

Energy Density of Electric Field

High

Energy Stored in Capacitor

High

Polar and Non-polar Dielectrics

Learning Objectives

Practice, Analyze & Improve 🚀

Revision Notes & Summary

Electrostatic Potential and Capacitance

Introduction

Key Concepts

Exercises

Important Formulas

Points to Ponder

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Practice Test – MCQs, True/False

Multiple Choice Questions

Q1.A uniformly charged spherical shell with charge QQQ and radius RRR is placed in an external electric field. What is the potential difference between the center of the shell and a point just outside the shell?

Correct Answer: D

Q2.A positive point charge QQQ is placed at the origin. What is the potential at a point PPP located at a distance rrr from the origin?

Correct Answer: C

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True or False

Q1.The potential at a point due to a system of point charges is calculated by vector addition of individual potentials.

Correct Answer: False

Q2.Inside a conductor, the electrostatic field is zero in the static situation.

Correct Answer: True

Q3.Equipotential surfaces are always perpendicular to electric field lines.

Correct Answer: True

High

Electrostatic Potential and Potential Difference

High

Parallel Plate Capacitor

High

Capacitors in Series

High

Dielectrics and Polarization

High

Electrostatics of Conductors

High

Electron Volt and Energy from Potential Difference

Medium

Potential Due to a System of Charges

Medium

Potential Energy of Dipole in External Electric Field

Medium

Capacitance and Capacitors

Medium

Dielectric Constant, Permittivity and Susceptibility

Medium

Electrostatic Potential Energy

Medium

Electric Potential Due to Point Charge

Medium

Capacitors in Parallel

Medium

Electric Field at Surface of Charged Conductor

Medium

Electrostatic Shielding and Cavities

Medium

Potential of Uniformly Charged Spherical Shell

Medium

Equipotential Surfaces

Medium

Relation Between Electric Field and Potential

Medium

Potential Energy of Charge Systems