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Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance is a fundamental chapter in electrostatics that covers electric potential, potential difference, capacitors, dielectrics, combinations of capacitors, and energy storage. These concepts form the foundation for current electricity and play a vital role in solving both conceptual and numerical problems in JEE Main and JEE Advanced.

ALLEN provides comprehensive theory for Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance, along with key formulas, solved examples, PYQ insights, and expert exam tips to help students revise efficiently, master the chapter, and strengthen their preparation for JEE Main, JEE Advanced, and CBSE Board exams.

1.0Chapter Snapshot

Parameter	Details
Difficulty Level	Moderate to High
JEE Main Weightage	High (Generally 1–2 Questions)
JEE Advanced Weightage	High (2–4 Integrated Concepts)
Key Concepts Covered	Electric Potential, Potential Difference, Equipotential Surfaces, Capacitance, Capacitors, Dielectrics, Energy Stored in Capacitors

2.0Chapter 2 Smart Study Material by ALLEN Experts: Download PDFs

Continue your preparation by exploring these resources after completing Electrostatic Potential and Capacitance. These study materials will help reinforce concepts, improve problem-solving speed, and strengthen exam readiness.

Practice Resources

Chapter Mock Test Paper
Chapter Mock Test Solutions
JEE Main PYQs with Solutions
JEE Advanced PYQs with Solutions
JEE Important Questions & Solutions on Electrostatics
NEET Important Questions & Solutions on Electrostatics

Test your Knowledge

question 1 of 5

The electrostatic potential at a point is defined as the work done in bringing a unit positive charge from:

1.The point to infinity

2.Infinity to that point

3.One point to another point

4.The origin to that point

Must-Revise Concepts

Capacitance

Capacitor and Capacitance

Capacitors

Capacitors and Capacitance

capacitors in combination

dielectric constant

effect dielectric capacitance

Electric Potential And Capacitance

Energy Stored in a capacitor

Equipotential Surfaces

Parallel Plate Capacitor

Parallel Plate Capacitor Questions

Relation Between Electric Field And Electric Potential

Supercapacitor

Electric Potential And Capacitance

Electric potential is the potential energy per unit charge in an electric field, influenced by point charges, dipoles, and charged spheres. Capacitance is a conductor's ability to store charge, depending on its geometry and surrounding dielectrics. Capacitors store energy and can be combined in series or parallel to adjust total capacitance.e

1.0Definition Electric Potential

It is the work done by an external force to move a unit positive charge from a reference point to a given point without changing its kinetic energy.

$V_{p} = \frac{( W _{\infty \to p} ) _{e x t}}{q}$ ( $Δ K = 0$ )

It is scalar quantity
Potential can be positive, negative and even zero
SI unit is joule/coulomb

2.0Electric Potential Due To a Point Charge

Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point against the electric field, without changing its kinetic energy.

Potential due to a positive point charge

$V_{P} = \frac{k Q}{r}$

Potential due to a negative point charge

$V_{P} = - \frac{k Q}{r}$

Note: Where reference potential is 0 at infinity.

1.0Electric Potential Due to System of Charges

$V = V_{1} + V_{2} + V_{3} + ........ V_{n}$

$V_{1} = \frac{1}{4 π ε _{0}} \frac{q _{1}}{r _{1 P}}$ , $V_{2} = \frac{1}{4 π ε _{0}} \frac{q _{2}}{r _{2 P}}$ , $V_{3} = \frac{1}{4 π ε _{0}} \frac{q _{3}}{r _{3 P}}$ , $V_{n} = \frac{1}{4 π ε _{0}} \frac{q _{n}}{r _{n P}}$

2.0Potential Due to an Electric Dipole

At axial point

Potential Due to an Electric Dipole At axial point

Electric potential due to +q charge, $V_{1} = \frac{k q}{r - l}$

Electric potential due to –q charge, $V_{2} = - \frac{k q}{r + l}$

Net electric potential $V = V_{1} + V_{2} = \frac{k q \times 2 l}{r ^{2} - l ^{2}}$

If $r >> l, V = \frac{k p}{r ^{2}}$

At equatorial point

Potential Due to an Electric Dipole At equatorial point

Electric potential at P due to +q charge, $V_{1} = \frac{k q}{x}$

Electric potential at P due to -q charge, $V_{1} = - \frac{k q}{x}$

Net electric potential $V = V_{1} + V_{2} = 0$

Potential at general point (r, $θ$ )

Potential Due to an Electric Dipole at general point

So, net potential at point A, $V = \frac{k ( p cos θ}{r ^{2}}$

$V = \frac{k ( p . r )}{r ^{3}} Here r = O A$

3.0Potential due to a Uniformly Charged Ring or Circular Arc

At the centre

Potential due to a Uniformly Charged Ring At the centre

$V = \frac{k Q}{R}$

On its axis

Potential due to a Uniformly Charged Ring On its axis

$V = \frac{Q}{4 π ε _{0} ( R ^{2} + ^{2} ) ^{\frac{1}{2}}} = \frac{k Q}{R ^{2} + x ^{2}} = \frac{k Q}{r}$

Potential due to a Uniformly Charged Circular Arc

Potential due to a Uniformly Charged Circular Arc

$V = \frac{k Q}{R} = \frac{k ( θR λ )}{R} = k θ λ$

4.0Potential Due to a Uniformly Charged Sphere

Electric Potential due to a Charged Conducting Sphere

Electric Potential due to a Charged Conducting Sphere

a. For outside points (r > R)

$V = \frac{k Q}{r} = \frac{σ R ^{2}}{ε _{0} r}$

b. For points on the surface (r = R)

$V = \frac{σ R}{ε _{0}} = \frac{k Q}{R}$

c. For Inside points (r < R)

$V = \frac{k Q}{r}$

Variation of potential with r

Electric Potential due to Uniformly Charged Non-conducting Sphere

Electric Potential due to Uniformly Charged Non-conducting Sphere

a. Outside (r > R)

$V = \frac{k Q}{r}$

b. On the surface (r = R)

$V = \frac{k Q}{r}$

c. Inside (r < R)

$V_{in s i d e} = \frac{k Q}{2 R ^{3}} [3 R^{2} - r^{2}]$

Variation of potential with r

5.0Relationship Between Electric Field and Electric Potential

$F = - \frac{d U}{d r}$

Dividing both sides by q

$\frac{F}{q} = - \frac{d}{d r} (\frac{U}{q}) \Rightarrow E = - \frac{d V}{d r} (F = qE an d U = q V)$

$E = - ▽ V$

$E_{x} = - \frac{\partial V}{\partial x}$ , $E_{y} = - \frac{\partial V}{\partial y}$ , $E_{z} = - \frac{\partial V}{\partial z}$

6.0Equipotential Surfaces

The locus of all such points which are at same potential or the surface at which potential of all the points are equal is known as equipotential surface (EPS). It can be of any shape.

Properties of equipotential surface

Electric field intensity is always perpendicular to EPS.
Work done in moving a charge from one point of EPS to another point of the same EPS is always zero.
Two EPS can never intersect each other.

EPS due to point charge - EPS of point charge is spherical in shape.

EPS due to line charge – EPS of line charge is cylindrical in shape.

EPS due to line charge – EPS of line charge is cylindrical in shape

EPS due to large plane charged sheet – EPS of a large plane sheet is plane in shape.

EPS due to large plane charged sheet – EPS of a large plane sheet is plane in shape.

EPS due to unlike charges -

EPS due to like charges -

7.0Electric Potential Energy of a System of Charges

Electrostatic potential energy is the work done to bring a charge from infinity to its position without changing its kinetic energy.

Electrostatic Potential Energy (EPE) of two point charge system

$Δ U = (W_{\infty \to r})_{e x t} = - \int_{\infty}^{r} F_{e x t} . d r$

$U_{r} - U_{in f t y} = - \int_{\infty}^{r} F_{e x t} . d r = \frac{k Qq}{r}$

$(∴ U_{\infty} = 0)$

$U - 0 = \frac{k Qq}{r}$

$U = \frac{k Qq}{r}$

8.0Electrostatic Potential energy of a three system of charge

$U_{S} = U_{12} + U_{23} + U_{13}$

$U_{S} = \frac{k q _{1} q _{2}}{r _{12}} + \frac{k q _{2} q _{3}}{r _{23}} + \frac{k q _{1} q _{3}}{r _{13}}$

Note: For n point charges system

No. of pairs= $\frac{n ( n - 1 )}{2}$

9.0Electrostatics of Conductors

Electric Field Inside a Conductor: The electric field inside a conductor in electrostatic equilibrium is zero.
Charge Distribution: Any excess charge on a conductor resides entirely on its surface.
Equipotential Surface: The entire conductor (including its surface) is at the same potential in electrostatic equilibrium.
Electric Field Just Outside the Surface: The field is perpendicular to the surface.
No Electric Field Parallel to the Surface: If there were, charges would move, contradicting equilibrium.
Cavity Inside a Conductor (No Charge):The electric field inside a cavity with no charge is zero, even if the conductor is charged.
Cavity with Internal Charge: Induced charges appear on the inner surface of the cavity to cancel the field inside the conductor.
Grounding: Connecting a conductor to the Earth allows excess charge to flow, maintaining zero potential.

10.0Properties of Conductors in Electrostatic Equilibrium

Conductors are materials which contain a large number of free electrons which can move freely inside the conductor.
Electrostatic field lines never exist inside a conductor.
Electric field lines terminate & originate always perpendicular to the surface of conductor because the surface of conductor is an equipotential surface.
Charge always resides on the outer surface of a conductor.
If there is a cavity inside a charged conductor with the cavity devoid of any charge then charge will always reside only on the outer surface of the conductor.
Electric field intensity near a conducting surface is given by, $E = \frac{σ}{ε _{0}} \overset{n}{^}$
When a conductor is grounded (earthed), its electric potential becomes zero. However, this doesn't always mean its charge is zero—only if the body is isolated can grounding ensure the charge is also zero.

11.0Electrostatic Shielding

It is the method of protecting a certain region from the effect of an electric field.
A cavity surrounded by conducting walls is a field free region as long as there are no charges inside the cavity.
Whatever be the charge and field configuration the field inside the cavity is always zero (Provided no charge present inside the cavity) This is known as electrostatic shielding.
Electrostatic shielding can be achieved by enclosing sensitive instruments in a hollow conductor.
Figure gives a summary of the important electrostatic properties of a conductor.

12.0Capacitance of a Conductor

It demonstrates a conductor's ability to store electrical energy through an electric field. If charge(Q) is given to an isolated conducting body and it's potential increases by V, then

$\Rightarrow$ Q ∝ V , Q =CV

$\Rightarrow= \frac{Q}{V}$ (C= Capacitance of the capacitor)

Electrical capacitance is a Scalar quantity.
Capacitance of conductor depends upon shape, size, presence of medium and nearness of other conductor.

Graph Between Q and V

S I Unit- Farad
CGS Unit-Stat Farad
Dimensional Formula-[M^-1L^-2T⁴A²]
Combination of Capacitors

13.0Combination of Capacitors

Series Combination

Capacitors are connected end-to-end so that the same current flows through each Capacitor. The total Capacitance in series is less than any individual capacitor's Capacitance. The Charge on each Capacitor connected in series is the same.
When two capacitors are connected in series, then effective capacitance is given by

$\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}}$

$\frac{1}{C} = \frac{C _{1} + C _{2}}{C _{1} C _{2}}$

$C = \frac{C _{1} C _{2}}{C _{1} + C _{2}}$

The effective capacitance of capacitors in series connection is lower than the capacitance of each capacitor individually.
The Charge for each capacitor in the series is the same.

Parallel Combination of Capacitor

Capacitors are connected across each other's terminals and share the same voltage. When a potential difference V is applied across the terminals all capacitors have equal potential difference. The equivalent Capacitance of parallel combination is more significant than any of the capacitances in the combination.
Effective Capacitance is given by, C=C₁+C₂+C₃
Effective Capacitance of parallel combination is greater than any of the capacitance.
In Parallel combination, voltage across each Capacitor is the same.

Parallel Plate Capacitor(PPC)

It comprises two large, flat, parallel conducting plates separated by a small distance.
$C = \frac{ε _{0} A}{d}$
Capacitance of parallel plate capacitor with dielectric

$C = \frac{ε _{0} ε _{r} A}{d} = ε_{r} C_{0} (∴ C_{0} = \frac{ε _{0} A}{d})$

Capacitance of Parallel Plate Capacitor Depends on

Area $\Rightarrow C \propto A$
Distance between the plates $\Rightarrow C \propto \frac{1}{d}$
Medium between the Plates $\Rightarrow C \propto ε_{r}$

Capacitance of Isolated Spherical Conductor

If the medium around the conductor is vacuum or air than capacitance is given by, $C = 4 π ϵ_{0} R$ (∴R = Radius of Spherical Conductor(Solid or Hollow)
If the medium around the conductor is a dielectric of constant K from surface of sphere to infinity then $C_{m e d i u m} = 4 π ε_{0} K R$
$\frac{C _{m e d i u m}}{C _{ai r / v a c uu m}} = K$ =Dielectric Constant

Capacitance of Spherical Capacitor

Case 1: Outer Sphere is Earthed

Capacitance of Spherical Capacitor when Outer Sphere is Earthed

When a charge Q is given to the inner sphere it is uniformly distributed on its surface A charge –Q is induced on the inner surface of the outer sphere. The charge +Q induced on the outer surface of the outer sphere flows to earth as it is grounded.
E=0 For r<a and E=0 for r>b ,this arrangement is known as spherical capacitor
Capacitance of spherical capacitor is given by $C = \frac{4 π ε _{0} ab}{b - a}$
If dielectric medium is filled then $C = \frac{4 π ε _{0} ε _{r} ab}{b - a}$

Case 2: Inner Sphere is earthed

Capacitance of Spherical Capacitor when Inner Sphere is earthed

Capacitance is given by, $C = \frac{4 π ε _{0} b ^{2}}{b - a}$

Cylindrical Capacitor

There are two coaxial conducting cylindrical surfaces where l >> a and l >> b, where a and b are the radius of cylinders. When a charge Q is given to the inner cylinder it is uniformly distributed on its surface. A charge –Q is induced on the inner surface of the outer cylinder. The charge +Q induced on outer surface of outer cylinder flows to earth as it is grounded. Hence Capacitance is given by $C = \frac{2 π ε _{0} l}{l n ( \frac{b}{a} )}$

14.0Dielectrics and Their Effect on Capacitance

Capacitors with Dielectric

Electric Field in the absence of Dielectric, $E = \frac{σ}{ε _{0}}$
Electric Field in the absence of Dielectric, $E_{0} = \frac{Q}{ε _{0} A}$
Electric Field in absence of Dielectric, $E = \frac{V}{d}$
Capacitance in absence of Dielectric, $C_{0} = \frac{Q}{V _{0}}$
Capacitance in presence of Dielectric, $C = \frac{Q - Q _{b}}{V}$
Dielectric Constant or Relative Permittivity (K or $ε_{r}$ )

$K = \frac{ϵ}{ϵ _{0}} = \frac{E _{0}}{E} = \frac{V _{0}}{V} = \frac{C}{C _{0}} = \frac{Q}{Q - Q _{b}} = \frac{σ}{σ - b}$

In presence of dielectric, capacitance is increased by a factor K

$C = \frac{ϵ _{0} A K}{d} = K C_{0}$

Capacitance of Parallel plate Capacitor when dielectric is partially filled

Capacitance is given by, $C = \frac{ϵ _{0} A}{( d - t ) + \frac{t}{ϵ _{r}}}$

15.0Effects of Dielectrics in Capacitor

Distance Division

Distance is Divided and area remains same
Capacitors are in Series
Individual Capacitance are $C_{1} = \frac{ϵ _{0} ϵ _{r 1} A}{d _{1}}$ and $C_{2} = \frac{ϵ _{0} ϵ _{r 2} A}{d _{2}}$

Effects of Dielectrics in Capacitor (Distance Division)

Effective Capacitance is given by, $C = ϵ_{0} A [\frac{ϵ _{r 1} ϵ _{r 2}}{d _{1} ϵ _{r 2} + d _{2} ϵ _{r 1}}]$
Special Case: If $d_{1} = d_{2} = \frac{d}{2} \Rightarrow C = \frac{ϵ _{0} A}{d} [\frac{2 ϵ _{r 1} ϵ _{r 2}}{ϵ _{r 1} + ϵ _{r 2}}]$

Area Division

Area is divided and distance remains same
Capacitors are in Parallel
Individual capacitors are

$C_{1} = \frac{ϵ _{0} ϵ _{r 1} A _{1}}{d}$ $C_{2} = \frac{ϵ _{0} ϵ _{r 2} A _{2}}{d}$

Effects of Dielectrics in Capacitor(Area Division)

Effective Capacitance is given by $C = \frac{ϵ _{0} ϵ _{r 1} A _{1}}{d} + \frac{ϵ _{0} ϵ _{r 2} A _{2}}{d}$
Special Case: If $A_{1} = A_{2} = \frac{A}{2} \Rightarrow C = \frac{ϵ _{0} A}{d} (\frac{ϵ _{r 1} + ϵ _{r 2}}{2})$

16.0Energy Stored in a Capacitor

A capacitor is a device designed to store electrical energy. The process of charging a capacitor entails transferring electric charges from one plate to another. The work done during this charging process is stored as electrical potential energy within the capacitor.

$V = \frac{W}{q} = \frac{d W}{d q}$

$d W = V d q = \frac{q}{C} d q$

$W = \int d W = \int_{0}^{Q} \frac{q}{C} d q = \frac{1}{C} \frac{Q ^{2}}{2} = \frac{Q ^{2}}{2 C}$

$U = \frac{Q ^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} Q V$

17.0Sample Questions on Electric Potential and Capacitance

Q-1. A conducting sphere of radius R has electric field E at distance 3R from the centre. Find potential difference between the centre of the sphere & the point where the electric field is E.

Solution:

$E = \frac{k Q}{( 3 R ) ^{2}} = \frac{k Q}{9 R ^{2}} \Rightarrow k Q = 9 E R^{2}$

$V_{C e n t re} = \frac{k Q}{R} = \frac{9 E R ^{2}}{R} = 9 ER$

$V_{P} = \frac{k Q}{r} = \frac{9 E R ^{2}}{3 R} = 3 ER$

$V_{C e n t re} - V_{P} = 9 ER - 3 ER = 6 ER$

Q-2. Find potential at points A,B and C due to the given arrangement of shells.

Solution:

$V_{A} = \frac{k Q _{1}}{a} + \frac{k Q _{2}}{b}$

$V_{B} = \frac{k Q _{1}}{a} + \frac{k Q _{2}}{b}$

$V_{C} = \frac{k Q _{1}}{b} + \frac{k Q _{2}}{b}$

Q-3. A charge in moving via three paths, 1, 2, and 3. as shown in figure. Find work done in each case.

Solution: Potential at points A, B, C and D is $V = \frac{k q}{R}$ . All the points are equipotential points, so work done in each case is zero.

Q-4. Calculate potential due to a charge $5 \times 1 0^{- 7}$ C at a point P located 9 cm away. Also find work done in bringing an another charge of $3 \times 1 0^{- 8}$ C from infinity to the point.

Solution:

Solved example questions on Electric Potential

$V_{P} = \frac{k q}{r} = \frac{9 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 7}}{9 \times 1 0 ^{- 2}} = 5 \times 1 0^{4} V$

$W = q (Δ V) = q (V_{P} - V_{in f t y}) = 3 \times 1 0^{- 8} (5 \times 1 0^{4} - 0) = 15 \times 1 0^{- 4} = 1.5 m J$

Q-5. Find p.d. between points A and B as shown in figure.

Problems on finding the potential difference between 2 points

Solution:

$V_{A} - V_{B} = 2 kλ l n (\frac{r _{B}}{r _{A}})$

$V_{A} - V_{B} = 2 kλ l n (2 a a)$

$V_{A} - V_{B} = 2 kλ l n (2)$

1.0Learning Outcomes

After completing Chapter 2: Electrostatic Potential and Capacitance, you will be able to:

Understand the concepts of electric potential and potential difference.
Calculate electric potential due to point charges, multiple charges, and electric dipoles.
Relate electric field intensity with electric potential using mathematical relationships.
Interpret equipotential surfaces and explain their important properties.
Calculate the capacitance of parallel plate capacitors and other capacitor configurations.
Solve problems involving series and parallel combinations of capacitors.
Understand the effect of dielectric materials on capacitance and electric fields.
Calculate the energy stored in capacitors under different conditions.
Strengthen problem-solving skills by mastering concepts and formulas frequently tested in JEE Main and JEE Advanced.

2.05-Minute Quick Revision

If you have only 5 minutes before your exam or mock test, focus on revising these high-weightage concepts and formulas to maximize your score.

Must-Revise Concepts

Important Formula Recall

Before attempting numerical questions, quickly revise these essential formulas:

Electric Potential due to a Point Charge
Electric Potential due to Multiple Charges
Electric Potential due to an Electric Dipole
Relation between Electric Field and Potential
Capacitance Formula
Parallel Plate Capacitor
Series Combination of Capacitors
Parallel Combination of Capacitors
Energy Stored in a Capacitor
Energy Density of Electric Field

3.0High Weightage Topics

The following topics have consistently appeared in JEE Main and JEE Advanced examinations and should be given higher priority during revision.

Topic	Focus Area
Electric Potential	Point charge, multiple charges
Potential Difference	Work done and conservative electric field
Equipotential Surfaces	Properties and field relationship
Potential due to Electric Dipole	Axial and equatorial positions
Capacitance	Parallel plate capacitor
Capacitor Combination	Series and parallel combinations
Dielectric Materials	Effect on capacitance and electric field
Energy Stored in Capacitors	Charging and discharging numericals

4.0Formula Handbook

The following formulas are the most important for solving JEE Main and JEE Advanced questions from Electrostatic Potential and Capacitance.

5.0Common Mistakes & JEE Tips

Electrostatic Potential and Capacitance is a concept-driven chapter where students often lose marks due to incorrect sign conventions, improper capacitor combinations, and confusion between electric potential and potential energy. Avoiding these mistakes can significantly improve your accuracy in JEE Main and JEE Advanced.

Common Mistake	Correct Approach
Confusing Electric Potential with Potential Energy	Electric Potential is energy per unit charge, whereas Potential Energy depends on the value of the charge.
Ignoring the sign of electric potential	Electric potential is an algebraic quantity; always consider the sign of the source charge.
Incorrect capacitor combinations	Use the correct formulas for series and parallel combinations.
Assuming capacitance depends on charge	Capacitance depends only on geometry and dielectric material.
Forgetting the effect of dielectrics	Dielectric materials increase capacitance and reduce the effective electric field.
Using incorrect energy formula	Select the appropriate energy expression based on the given variables.
Mixing electric field and electric potential concepts	Electric field is the negative gradient of electric potential.
Incorrect SI unit conversion	Always use SI units such as Volt (V), Farad (F), Coulomb (C), and Joule (J).

ALLEN Faculty Tips

Draw neat capacitor diagrams before solving numerical problems.
Memorize standard formulas for series and parallel capacitor combinations.
Understand the physical meaning of electric potential instead of memorizing formulas blindly.
Revise dielectric concepts thoroughly as they are frequently asked in JEE.
Practice previous years' questions on capacitor combinations and energy stored.
Revise important formulas regularly to improve speed and accuracy.

6.0PYQ Trend Analysis

Electrostatic Potential and Capacitance is one of the highest-weightage chapters in Class 12 Physics for both JEE Main and JEE Advanced. Questions are primarily numerical and concept-based, with a strong emphasis on electric potential calculations, capacitor combinations, dielectric materials, and energy stored in capacitors.

Note: The trend below is based on the analysis of previous years' JEE Main and JEE Advanced question papers. The exact number of questions may vary each year.

Concept	Questions Asked (Last 5 Years)	Average Difficulty
Electric Potential	5–6	Easy–Moderate
Potential Difference	3–4	Moderate
Equipotential Surfaces	2–3	Easy
Capacitance	7–8	Moderate
Capacitor Combinations	5–6	Moderate
Dielectric Materials	3–4	Moderate
Energy Stored in Capacitors	5–6	Moderate

Most Common Question Types

Electric potential due to point charges and charge distributions.
Work done in moving a charge between two points.
Relation between electric field and electric potential.
Series and parallel capacitor combinations.
Dielectric insertion problems.
Energy stored in capacitors.
Assertion–Reason and concept-based MCQs.

Smart Revision Strategy

If you're short on time, revise the chapter in the following order:

Electric Potential
Potential Difference
Equipotential Surfaces
Relation between Electric Field & Potential
Capacitance
Parallel Plate Capacitor
Capacitor Combinations
Dielectric Materials
Energy Stored in Capacitors

Class 12 Physics Chapters List:- JEE & NEET Revision Notes

Chapter 1 Electric Charges and Fields

Chapter 2 Electrostatic Potential and Capacitance

Chapter 3 Current Electricity

Chapter 4 Moving Charges and Magnetism

Chapter 5 Magnetism and Matter

Chapter 6 Electromagnetic Induction

Chapter 7 Alternating Current

Chapter 8 Electromagnetic Waves

Chapter 9 Ray Optics and Optical Instruments

Chapter 10 Wave Optics

Chapter 11 Dual Nature of Radiation and Matter

Chapter 12 Atoms

Chapter 13 Nuclei

Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

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Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance is a fundamental chapter in electrostatics that covers electric potential, potential difference, capacitors, dielectrics, combinations of capacitors, and energy storage. These concepts form the foundation for current electricity and play a vital role in solving both conceptual and numerical problems in JEE Main and JEE Advanced.

ALLEN provides comprehensive theory for Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance, along with key formulas, solved examples, PYQ insights, and expert exam tips to help students revise efficiently, master the chapter, and strengthen their preparation for JEE Main, JEE Advanced, and CBSE Board exams.

1.0Chapter Snapshot

Parameter	Details
Difficulty Level	Moderate to High
JEE Main Weightage	High (Generally 1–2 Questions)
JEE Advanced Weightage	High (2–4 Integrated Concepts)
Key Concepts Covered	Electric Potential, Potential Difference, Equipotential Surfaces, Capacitance, Capacitors, Dielectrics, Energy Stored in Capacitors

2.0Chapter 2 Smart Study Material by ALLEN Experts: Download PDFs

Continue your preparation by exploring these resources after completing Electrostatic Potential and Capacitance. These study materials will help reinforce concepts, improve problem-solving speed, and strengthen exam readiness.

Practice Resources

Chapter Mock Test Paper
Chapter Mock Test Solutions
JEE Main PYQs with Solutions
JEE Advanced PYQs with Solutions
JEE Important Questions & Solutions on Electrostatics
NEET Important Questions & Solutions on Electrostatics

Electric Potential And Capacitance

Electric potential is the potential energy per unit charge in an electric field, influenced by point charges, dipoles, and charged spheres. Capacitance is a conductor's ability to store charge, depending on its geometry and surrounding dielectrics. Capacitors store energy and can be combined in series or parallel to adjust total capacitance.e

1.0Definition Electric Potential

It is the work done by an external force to move a unit positive charge from a reference point to a given point without changing its kinetic energy.

$V_{p} = \frac{( W _{\infty \to p} ) _{e x t}}{q}$ ( $Δ K = 0$ )

It is scalar quantity
Potential can be positive, negative and even zero
SI unit is joule/coulomb

2.0Electric Potential Due To a Point Charge

Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point against the electric field, without changing its kinetic energy.

Potential due to a positive point charge

$V_{P} = \frac{k Q}{r}$

Potential due to a negative point charge

$V_{P} = - \frac{k Q}{r}$

Note: Where reference potential is 0 at infinity.

1.0Electric Potential Due to System of Charges

$V = V_{1} + V_{2} + V_{3} + ........ V_{n}$

$V_{1} = \frac{1}{4 π ε _{0}} \frac{q _{1}}{r _{1 P}}$ , $V_{2} = \frac{1}{4 π ε _{0}} \frac{q _{2}}{r _{2 P}}$ , $V_{3} = \frac{1}{4 π ε _{0}} \frac{q _{3}}{r _{3 P}}$ , $V_{n} = \frac{1}{4 π ε _{0}} \frac{q _{n}}{r _{n P}}$

2.0Potential Due to an Electric Dipole

At axial point

Electric potential due to +q charge, $V_{1} = \frac{k q}{r - l}$

Electric potential due to –q charge, $V_{2} = - \frac{k q}{r + l}$

Net electric potential $V = V_{1} + V_{2} = \frac{k q \times 2 l}{r ^{2} - l ^{2}}$

If $r >> l, V = \frac{k p}{r ^{2}}$

At equatorial point

Electric potential at P due to +q charge, $V_{1} = \frac{k q}{x}$

Electric potential at P due to -q charge, $V_{1} = - \frac{k q}{x}$

Net electric potential $V = V_{1} + V_{2} = 0$

Potential at general point (r, $θ$ )

So, net potential at point A, $V = \frac{k ( p cos θ}{r ^{2}}$

$V = \frac{k ( p . r )}{r ^{3}} Here r = O A$

3.0Potential due to a Uniformly Charged Ring or Circular Arc

At the centre

$V = \frac{k Q}{R}$

On its axis

$V = \frac{Q}{4 π ε _{0} ( R ^{2} + ^{2} ) ^{\frac{1}{2}}} = \frac{k Q}{R ^{2} + x ^{2}} = \frac{k Q}{r}$

Potential due to a Uniformly Charged Circular Arc

$V = \frac{k Q}{R} = \frac{k ( θR λ )}{R} = k θ λ$

4.0Potential Due to a Uniformly Charged Sphere

Electric Potential due to a Charged Conducting Sphere

a. For outside points (r > R)

$V = \frac{k Q}{r} = \frac{σ R ^{2}}{ε _{0} r}$

b. For points on the surface (r = R)

$V = \frac{σ R}{ε _{0}} = \frac{k Q}{R}$

c. For Inside points (r < R)

$V = \frac{k Q}{r}$

Variation of potential with r

Electric Potential due to Uniformly Charged Non-conducting Sphere

a. Outside (r > R)

$V = \frac{k Q}{r}$

b. On the surface (r = R)

$V = \frac{k Q}{r}$

c. Inside (r < R)

$V_{in s i d e} = \frac{k Q}{2 R ^{3}} [3 R^{2} - r^{2}]$

Variation of potential with r

5.0Relationship Between Electric Field and Electric Potential

$F = - \frac{d U}{d r}$

Dividing both sides by q

$\frac{F}{q} = - \frac{d}{d r} (\frac{U}{q}) \Rightarrow E = - \frac{d V}{d r} (F = qE an d U = q V)$

$E = - ▽ V$

$E_{x} = - \frac{\partial V}{\partial x}$ , $E_{y} = - \frac{\partial V}{\partial y}$ , $E_{z} = - \frac{\partial V}{\partial z}$

6.0Equipotential Surfaces

The locus of all such points which are at same potential or the surface at which potential of all the points are equal is known as equipotential surface (EPS). It can be of any shape.

Properties of equipotential surface

Electric field intensity is always perpendicular to EPS.
Work done in moving a charge from one point of EPS to another point of the same EPS is always zero.
Two EPS can never intersect each other.

EPS due to point charge - EPS of point charge is spherical in shape.

EPS due to line charge – EPS of line charge is cylindrical in shape.

EPS due to large plane charged sheet – EPS of a large plane sheet is plane in shape.

EPS due to unlike charges -

EPS due to like charges -

7.0Electric Potential Energy of a System of Charges

Electrostatic potential energy is the work done to bring a charge from infinity to its position without changing its kinetic energy.

Electrostatic Potential Energy (EPE) of two point charge system

$Δ U = (W_{\infty \to r})_{e x t} = - \int_{\infty}^{r} F_{e x t} . d r$

$U_{r} - U_{in f t y} = - \int_{\infty}^{r} F_{e x t} . d r = \frac{k Qq}{r}$

$(∴ U_{\infty} = 0)$

$U - 0 = \frac{k Qq}{r}$

$U = \frac{k Qq}{r}$

8.0Electrostatic Potential energy of a three system of charge

$U_{S} = U_{12} + U_{23} + U_{13}$

$U_{S} = \frac{k q _{1} q _{2}}{r _{12}} + \frac{k q _{2} q _{3}}{r _{23}} + \frac{k q _{1} q _{3}}{r _{13}}$

Note: For n point charges system

No. of pairs= $\frac{n ( n - 1 )}{2}$

9.0Electrostatics of Conductors

Electric Field Inside a Conductor: The electric field inside a conductor in electrostatic equilibrium is zero.
Charge Distribution: Any excess charge on a conductor resides entirely on its surface.
Equipotential Surface: The entire conductor (including its surface) is at the same potential in electrostatic equilibrium.
Electric Field Just Outside the Surface: The field is perpendicular to the surface.
No Electric Field Parallel to the Surface: If there were, charges would move, contradicting equilibrium.
Cavity Inside a Conductor (No Charge):The electric field inside a cavity with no charge is zero, even if the conductor is charged.
Cavity with Internal Charge: Induced charges appear on the inner surface of the cavity to cancel the field inside the conductor.
Grounding: Connecting a conductor to the Earth allows excess charge to flow, maintaining zero potential.

10.0Properties of Conductors in Electrostatic Equilibrium

Conductors are materials which contain a large number of free electrons which can move freely inside the conductor.
Electrostatic field lines never exist inside a conductor.
Electric field lines terminate & originate always perpendicular to the surface of conductor because the surface of conductor is an equipotential surface.
Charge always resides on the outer surface of a conductor.
If there is a cavity inside a charged conductor with the cavity devoid of any charge then charge will always reside only on the outer surface of the conductor.
Electric field intensity near a conducting surface is given by, $E = \frac{σ}{ε _{0}} \overset{n}{^}$
When a conductor is grounded (earthed), its electric potential becomes zero. However, this doesn't always mean its charge is zero—only if the body is isolated can grounding ensure the charge is also zero.

11.0Electrostatic Shielding

It is the method of protecting a certain region from the effect of an electric field.
A cavity surrounded by conducting walls is a field free region as long as there are no charges inside the cavity.
Whatever be the charge and field configuration the field inside the cavity is always zero (Provided no charge present inside the cavity) This is known as electrostatic shielding.
Electrostatic shielding can be achieved by enclosing sensitive instruments in a hollow conductor.
Figure gives a summary of the important electrostatic properties of a conductor.

12.0Capacitance of a Conductor

It demonstrates a conductor's ability to store electrical energy through an electric field. If charge(Q) is given to an isolated conducting body and it's potential increases by V, then

$\Rightarrow$ Q ∝ V , Q =CV

$\Rightarrow= \frac{Q}{V}$ (C= Capacitance of the capacitor)

Electrical capacitance is a Scalar quantity.
Capacitance of conductor depends upon shape, size, presence of medium and nearness of other conductor.

Graph Between Q and V

S I Unit- Farad
CGS Unit-Stat Farad
Dimensional Formula-[M^-1L^-2T⁴A²]
Combination of Capacitors

13.0Combination of Capacitors

Series Combination

Capacitors are connected end-to-end so that the same current flows through each Capacitor. The total Capacitance in series is less than any individual capacitor's Capacitance. The Charge on each Capacitor connected in series is the same.
When two capacitors are connected in series, then effective capacitance is given by

$\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}}$

$\frac{1}{C} = \frac{C _{1} + C _{2}}{C _{1} C _{2}}$

$C = \frac{C _{1} C _{2}}{C _{1} + C _{2}}$

The effective capacitance of capacitors in series connection is lower than the capacitance of each capacitor individually.
The Charge for each capacitor in the series is the same.

Parallel Combination of Capacitor

Capacitors are connected across each other's terminals and share the same voltage. When a potential difference V is applied across the terminals all capacitors have equal potential difference. The equivalent Capacitance of parallel combination is more significant than any of the capacitances in the combination.
Effective Capacitance is given by, C=C₁+C₂+C₃
Effective Capacitance of parallel combination is greater than any of the capacitance.
In Parallel combination, voltage across each Capacitor is the same.

Parallel Plate Capacitor(PPC)

It comprises two large, flat, parallel conducting plates separated by a small distance.
$C = \frac{ε _{0} A}{d}$
Capacitance of parallel plate capacitor with dielectric

$C = \frac{ε _{0} ε _{r} A}{d} = ε_{r} C_{0} (∴ C_{0} = \frac{ε _{0} A}{d})$

Capacitance of Parallel Plate Capacitor Depends on

Area $\Rightarrow C \propto A$
Distance between the plates $\Rightarrow C \propto \frac{1}{d}$
Medium between the Plates $\Rightarrow C \propto ε_{r}$

Capacitance of Isolated Spherical Conductor

If the medium around the conductor is vacuum or air than capacitance is given by, $C = 4 π ϵ_{0} R$ (∴R = Radius of Spherical Conductor(Solid or Hollow)
If the medium around the conductor is a dielectric of constant K from surface of sphere to infinity then $C_{m e d i u m} = 4 π ε_{0} K R$
$\frac{C _{m e d i u m}}{C _{ai r / v a c uu m}} = K$ =Dielectric Constant

Capacitance of Spherical Capacitor

Case 1: Outer Sphere is Earthed

When a charge Q is given to the inner sphere it is uniformly distributed on its surface A charge –Q is induced on the inner surface of the outer sphere. The charge +Q induced on the outer surface of the outer sphere flows to earth as it is grounded.
E=0 For r<a and E=0 for r>b ,this arrangement is known as spherical capacitor
Capacitance of spherical capacitor is given by $C = \frac{4 π ε _{0} ab}{b - a}$
If dielectric medium is filled then $C = \frac{4 π ε _{0} ε _{r} ab}{b - a}$

Case 2: Inner Sphere is earthed

Capacitance is given by, $C = \frac{4 π ε _{0} b ^{2}}{b - a}$

Cylindrical Capacitor

There are two coaxial conducting cylindrical surfaces where l >> a and l >> b, where a and b are the radius of cylinders. When a charge Q is given to the inner cylinder it is uniformly distributed on its surface. A charge –Q is induced on the inner surface of the outer cylinder. The charge +Q induced on outer surface of outer cylinder flows to earth as it is grounded. Hence Capacitance is given by $C = \frac{2 π ε _{0} l}{l n ( \frac{b}{a} )}$

14.0Dielectrics and Their Effect on Capacitance

Capacitors with Dielectric

Electric Field in the absence of Dielectric, $E = \frac{σ}{ε _{0}}$
Electric Field in the absence of Dielectric, $E_{0} = \frac{Q}{ε _{0} A}$
Electric Field in absence of Dielectric, $E = \frac{V}{d}$
Capacitance in absence of Dielectric, $C_{0} = \frac{Q}{V _{0}}$
Capacitance in presence of Dielectric, $C = \frac{Q - Q _{b}}{V}$
Dielectric Constant or Relative Permittivity (K or $ε_{r}$ )

$K = \frac{ϵ}{ϵ _{0}} = \frac{E _{0}}{E} = \frac{V _{0}}{V} = \frac{C}{C _{0}} = \frac{Q}{Q - Q _{b}} = \frac{σ}{σ - b}$

In presence of dielectric, capacitance is increased by a factor K

$C = \frac{ϵ _{0} A K}{d} = K C_{0}$

Capacitance of Parallel plate Capacitor when dielectric is partially filled

Capacitance is given by, $C = \frac{ϵ _{0} A}{( d - t ) + \frac{t}{ϵ _{r}}}$

15.0Effects of Dielectrics in Capacitor

Distance Division

Distance is Divided and area remains same
Capacitors are in Series
Individual Capacitance are $C_{1} = \frac{ϵ _{0} ϵ _{r 1} A}{d _{1}}$ and $C_{2} = \frac{ϵ _{0} ϵ _{r 2} A}{d _{2}}$

Effective Capacitance is given by, $C = ϵ_{0} A [\frac{ϵ _{r 1} ϵ _{r 2}}{d _{1} ϵ _{r 2} + d _{2} ϵ _{r 1}}]$
Special Case: If $d_{1} = d_{2} = \frac{d}{2} \Rightarrow C = \frac{ϵ _{0} A}{d} [\frac{2 ϵ _{r 1} ϵ _{r 2}}{ϵ _{r 1} + ϵ _{r 2}}]$

Area Division

Area is divided and distance remains same
Capacitors are in Parallel
Individual capacitors are

$C_{1} = \frac{ϵ _{0} ϵ _{r 1} A _{1}}{d}$ $C_{2} = \frac{ϵ _{0} ϵ _{r 2} A _{2}}{d}$

Effective Capacitance is given by $C = \frac{ϵ _{0} ϵ _{r 1} A _{1}}{d} + \frac{ϵ _{0} ϵ _{r 2} A _{2}}{d}$
Special Case: If $A_{1} = A_{2} = \frac{A}{2} \Rightarrow C = \frac{ϵ _{0} A}{d} (\frac{ϵ _{r 1} + ϵ _{r 2}}{2})$

16.0Energy Stored in a Capacitor

A capacitor is a device designed to store electrical energy. The process of charging a capacitor entails transferring electric charges from one plate to another. The work done during this charging process is stored as electrical potential energy within the capacitor.

$V = \frac{W}{q} = \frac{d W}{d q}$

$d W = V d q = \frac{q}{C} d q$

$W = \int d W = \int_{0}^{Q} \frac{q}{C} d q = \frac{1}{C} \frac{Q ^{2}}{2} = \frac{Q ^{2}}{2 C}$

$U = \frac{Q ^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} Q V$

17.0Sample Questions on Electric Potential and Capacitance

Q-1. A conducting sphere of radius R has electric field E at distance 3R from the centre. Find potential difference between the centre of the sphere & the point where the electric field is E.

Solution:

$E = \frac{k Q}{( 3 R ) ^{2}} = \frac{k Q}{9 R ^{2}} \Rightarrow k Q = 9 E R^{2}$

$V_{C e n t re} = \frac{k Q}{R} = \frac{9 E R ^{2}}{R} = 9 ER$

$V_{P} = \frac{k Q}{r} = \frac{9 E R ^{2}}{3 R} = 3 ER$

$V_{C e n t re} - V_{P} = 9 ER - 3 ER = 6 ER$

Q-2. Find potential at points A,B and C due to the given arrangement of shells.

Solution:

$V_{A} = \frac{k Q _{1}}{a} + \frac{k Q _{2}}{b}$

$V_{B} = \frac{k Q _{1}}{a} + \frac{k Q _{2}}{b}$

$V_{C} = \frac{k Q _{1}}{b} + \frac{k Q _{2}}{b}$

Q-3. A charge in moving via three paths, 1, 2, and 3. as shown in figure. Find work done in each case.

Solution: Potential at points A, B, C and D is $V = \frac{k q}{R}$ . All the points are equipotential points, so work done in each case is zero.

Q-4. Calculate potential due to a charge $5 \times 1 0^{- 7}$ C at a point P located 9 cm away. Also find work done in bringing an another charge of $3 \times 1 0^{- 8}$ C from infinity to the point.

Solution:

$V_{P} = \frac{k q}{r} = \frac{9 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 7}}{9 \times 1 0 ^{- 2}} = 5 \times 1 0^{4} V$

$W = q (Δ V) = q (V_{P} - V_{in f t y}) = 3 \times 1 0^{- 8} (5 \times 1 0^{4} - 0) = 15 \times 1 0^{- 4} = 1.5 m J$

Q-5. Find p.d. between points A and B as shown in figure.

Solution:

$V_{A} - V_{B} = 2 kλ l n (\frac{r _{B}}{r _{A}})$

$V_{A} - V_{B} = 2 kλ l n (2 a a)$

$V_{A} - V_{B} = 2 kλ l n (2)$

1.0Learning Outcomes

After completing Chapter 2: Electrostatic Potential and Capacitance, you will be able to:

Understand the concepts of electric potential and potential difference.
Calculate electric potential due to point charges, multiple charges, and electric dipoles.
Relate electric field intensity with electric potential using mathematical relationships.
Interpret equipotential surfaces and explain their important properties.
Calculate the capacitance of parallel plate capacitors and other capacitor configurations.
Solve problems involving series and parallel combinations of capacitors.
Understand the effect of dielectric materials on capacitance and electric fields.
Calculate the energy stored in capacitors under different conditions.
Strengthen problem-solving skills by mastering concepts and formulas frequently tested in JEE Main and JEE Advanced.

2.05-Minute Quick Revision

If you have only 5 minutes before your exam or mock test, focus on revising these high-weightage concepts and formulas to maximize your score.

Must-Revise Concepts

Important Formula Recall

Before attempting numerical questions, quickly revise these essential formulas:

Electric Potential due to a Point Charge
Electric Potential due to Multiple Charges
Electric Potential due to an Electric Dipole
Relation between Electric Field and Potential
Capacitance Formula
Parallel Plate Capacitor
Series Combination of Capacitors
Parallel Combination of Capacitors
Energy Stored in a Capacitor
Energy Density of Electric Field

3.0High Weightage Topics

The following topics have consistently appeared in JEE Main and JEE Advanced examinations and should be given higher priority during revision.

Topic	Focus Area
Electric Potential	Point charge, multiple charges
Potential Difference	Work done and conservative electric field
Equipotential Surfaces	Properties and field relationship
Potential due to Electric Dipole	Axial and equatorial positions
Capacitance	Parallel plate capacitor
Capacitor Combination	Series and parallel combinations
Dielectric Materials	Effect on capacitance and electric field
Energy Stored in Capacitors	Charging and discharging numericals

4.0Formula Handbook

The following formulas are the most important for solving JEE Main and JEE Advanced questions from Electrostatic Potential and Capacitance.

5.0Common Mistakes & JEE Tips

Electrostatic Potential and Capacitance is a concept-driven chapter where students often lose marks due to incorrect sign conventions, improper capacitor combinations, and confusion between electric potential and potential energy. Avoiding these mistakes can significantly improve your accuracy in JEE Main and JEE Advanced.

Common Mistake	Correct Approach
Confusing Electric Potential with Potential Energy	Electric Potential is energy per unit charge, whereas Potential Energy depends on the value of the charge.
Ignoring the sign of electric potential	Electric potential is an algebraic quantity; always consider the sign of the source charge.
Incorrect capacitor combinations	Use the correct formulas for series and parallel combinations.
Assuming capacitance depends on charge	Capacitance depends only on geometry and dielectric material.
Forgetting the effect of dielectrics	Dielectric materials increase capacitance and reduce the effective electric field.
Using incorrect energy formula	Select the appropriate energy expression based on the given variables.
Mixing electric field and electric potential concepts	Electric field is the negative gradient of electric potential.
Incorrect SI unit conversion	Always use SI units such as Volt (V), Farad (F), Coulomb (C), and Joule (J).

ALLEN Faculty Tips

Draw neat capacitor diagrams before solving numerical problems.
Memorize standard formulas for series and parallel capacitor combinations.
Understand the physical meaning of electric potential instead of memorizing formulas blindly.
Revise dielectric concepts thoroughly as they are frequently asked in JEE.
Practice previous years' questions on capacitor combinations and energy stored.
Revise important formulas regularly to improve speed and accuracy.

6.0PYQ Trend Analysis

Electrostatic Potential and Capacitance is one of the highest-weightage chapters in Class 12 Physics for both JEE Main and JEE Advanced. Questions are primarily numerical and concept-based, with a strong emphasis on electric potential calculations, capacitor combinations, dielectric materials, and energy stored in capacitors.

Note: The trend below is based on the analysis of previous years' JEE Main and JEE Advanced question papers. The exact number of questions may vary each year.

Concept	Questions Asked (Last 5 Years)	Average Difficulty
Electric Potential	5–6	Easy–Moderate
Potential Difference	3–4	Moderate
Equipotential Surfaces	2–3	Easy
Capacitance	7–8	Moderate
Capacitor Combinations	5–6	Moderate
Dielectric Materials	3–4	Moderate
Energy Stored in Capacitors	5–6	Moderate

Most Common Question Types

Electric potential due to point charges and charge distributions.
Work done in moving a charge between two points.
Relation between electric field and electric potential.
Series and parallel capacitor combinations.
Dielectric insertion problems.
Energy stored in capacitors.
Assertion–Reason and concept-based MCQs.

Smart Revision Strategy

If you're short on time, revise the chapter in the following order:

Electric Potential
Potential Difference
Equipotential Surfaces
Relation between Electric Field & Potential
Capacitance
Parallel Plate Capacitor
Capacitor Combinations
Dielectric Materials
Energy Stored in Capacitors

Test your Knowledge

question 1 of 5

The electrostatic potential at a point is defined as the work done in bringing a unit positive charge from:

Frequently Asked Questions

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Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance

1.0Chapter Snapshot

2.0Chapter 2 Smart Study Material by ALLEN Experts: Download PDFs

Practice Resources

Test your Knowledge

question 1 of 5

The electrostatic potential at a point is defined as the work done in bringing a unit positive charge from:

Electric Potential And Capacitance

1.0Definition Electric Potential

2.0Electric Potential Due To a Point Charge

1.0Electric Potential Due to System of Charges

2.0Potential Due to an Electric Dipole

3.0Potential due to a Uniformly Charged Ring or Circular Arc

4.0Potential Due to a Uniformly Charged Sphere

5.0Relationship Between Electric Field and Electric Potential

6.0Equipotential Surfaces

7.0Electric Potential Energy of a System of Charges

8.0Electrostatic Potential energy of a three system of charge

9.0Electrostatics of Conductors

10.0Properties of Conductors in Electrostatic Equilibrium

11.0Electrostatic Shielding

12.0Capacitance of a Conductor

Graph Between Q and V

13.0Combination of Capacitors

Parallel Plate Capacitor(PPC)

Capacitance of Isolated Spherical Conductor

Capacitance of Spherical Capacitor

Cylindrical Capacitor

14.0Dielectrics and Their Effect on Capacitance

Capacitors with Dielectric

Capacitance of Parallel plate Capacitor when dielectric is partially filled

15.0Effects of Dielectrics in Capacitor

16.0Energy Stored in a Capacitor

17.0Sample Questions on Electric Potential and Capacitance

1.0Learning Outcomes

2.05-Minute Quick Revision

Must-Revise Concepts

Important Formula Recall

3.0High Weightage Topics

4.0Formula Handbook

Related Videos

5.0Common Mistakes & JEE Tips

ALLEN Faculty Tips

6.0PYQ Trend Analysis

Most Common Question Types

Smart Revision Strategy

Frequently Asked Questions

What are the most important topics in Electrostatic Potential and Capacitance?

Which formulas are most important in this chapter?

Is this chapter Electrostatic Potential and Capacitance numerical?

How many JEE questions are asked from this chapter?

Can these notes help with quick revision?

Join ALLEN!

Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance

1.0Chapter Snapshot

2.0Chapter 2 Smart Study Material by ALLEN Experts: Download PDFs

Practice Resources

Electric Potential And Capacitance

1.0Definition Electric Potential

2.0Electric Potential Due To a Point Charge

1.0Electric Potential Due to System of Charges

2.0Potential Due to an Electric Dipole

3.0Potential due to a Uniformly Charged Ring or Circular Arc

4.0Potential Due to a Uniformly Charged Sphere

5.0Relationship Between Electric Field and Electric Potential

6.0Equipotential Surfaces

7.0Electric Potential Energy of a System of Charges

8.0Electrostatic Potential energy of a three system of charge

9.0Electrostatics of Conductors

10.0Properties of Conductors in Electrostatic Equilibrium