Electrostatic notes

Electrostatic Notes
Unit I — Electrostatics
Class 12 Physics

Complete Unit I notes covering Electric Charges and Fields and Electrostatic Potential and Capacitance. The structure follows the supplied master syllabus, with detailed theory, step-by-step derivations, diagrams, examples, formula revision and board-answer guidance.

Class 12 Physics Unit 1 — Electrostatics
These free Class 12 Physics Electrostatics notes cover both major chapters: Electric Charges and Fields and Electrostatic Potential and Capacitance. Use the notes for concept building, board-exam derivations, numerical practice, formula revision and quick exam preparation.

More free Class 12 study material at UniversityScope.com →

Unit I Coverage
Chapter 1: Electric Charges and Fields • Chapter 2: Electrostatic Potential and Capacitance

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Unit roadmap

1. Chapter 1 — Electric Charges and Fields
2. Chapter 2 — Electrostatic Potential and Capacitance
3. Unit I Solved Numericals
4. Master Revision, Formula Sheet and Exam Strategy

Chapter 1 — Electric Charges and Fields

Charge is the source of electric interaction. This chapter develops the ideas of Coulomb force, electric field, field lines, electric dipole, electric flux and Gauss's law.

This chapter is the foundation of electrostatics. We will learn what electric charge is, how charges interact, what an electric field means, how an electric dipole behaves, what electric flux is, and how Gauss's law is used to calculate electric fields. All important derivations are written step by step in very simple English.

Chapter contents

1. Electric Charge and Conservation of Charge
2. Coulomb's Law
3. Multiple Charges and Superposition Principle
4. Continuous Charge Distribution
5. Electric Field and Field Due to a Point Charge
6. Electric Field Lines
7. Electric Dipole and Dipole Moment
8. Electric Field Due to a Dipole
9. Torque on a Dipole in a Uniform Electric Field
10. Electric Flux
11. Gauss's Theorem
12. Infinite Straight Wire
13. Infinite Plane Sheet
14. Uniformly Charged Thin Spherical Shell
15. Formula Sheet and Exam Points

1. Electric Charges

Electric charge is a basic property of matter. Because of electric charge, an object can experience an electric force. There are two types of charge:

• Positive charge
• Negative charge

Situation	Result
Positive + positive	Repulsion
Negative + negative	Repulsion
Positive + negative	Attraction

1.1 SI unit of charge

The SI unit of electric charge is coulomb (C).

1.2 Additivity of charge

If a body has many charges, its total charge is the algebraic sum of all the charges.

\[ \boxed{Q=q_1+q_2+q_3+\cdots+q_n} \]

1.3 Conservation of charge

Charge can move from one object to another, but charge cannot be created or destroyed in an isolated system.

\[ \boxed{Q_{\text{initial}}=Q_{\text{final}}} \]

For example, when a glass rod becomes positively charged by rubbing, electrons are transferred away from the rod. The total charge of the complete isolated system remains unchanged.

1.4 Quantisation of charge

At the microscopic level, charge occurs in integral multiples of the elementary charge.

\[ \boxed{q=ne} \]

where \(n=0,\pm1,\pm2,\ldots\), and \(e=1.602\times10^{-19}\,\mathrm C\).

Simple memory: Charge is additive, conserved and quantised.

2. Coulomb's Law

Coulomb's law gives the electrostatic force between two stationary point charges. Suppose two point charges \(q_1\) and \(q_2\) are separated by distance \(r\).

For like charges, the forces are repulsive.

2.1 Derivation of Coulomb's law

Step 1: Dependence on the first charge.
If \(q_2\) and \(r\) remain fixed, increasing \(q_1\) increases the force. Therefore, \[ F\propto q_1. \]

Step 2: Dependence on the second charge.
If \(q_1\) and \(r\) remain fixed, \[ F\propto q_2. \]

Step 3: Combine both.
\[ F\propto q_1q_2. \] For magnitude, we use \(|q_1q_2|\).

Step 4: Dependence on distance.
Experiment shows that electrostatic force follows the inverse-square law: \[ F\propto\frac1{r^2}. \]

Step 5: Combine all dependences.
\[ F\propto\frac{|q_1q_2|}{r^2}. \]

Step 6: Introduce the proportionality constant.
\[ F=k\frac{|q_1q_2|}{r^2}. \]

Step 7: Value of \(k\) in vacuum.
\[ k=\frac1{4\pi\varepsilon_0}. \]

Final result:
\[ \boxed{ F=\frac1{4\pi\varepsilon_0}\frac{|q_1q_2|}{r^2} } \]

Here \(\varepsilon_0\) is the permittivity of free space:

\[ \varepsilon_0=8.854\times10^{-12}\ \mathrm{C^2N^{-1}m^{-2}}. \]

2.2 Direction of force

• Like charges repel.
• Unlike charges attract.
• The force acts along the straight line joining the two charges.

2.3 Vector form

\[ \boxed{ \vec F_{12} = \frac1{4\pi\varepsilon_0} \frac{q_1q_2}{r^2}\hat r_{12} } \]

Exam point: If distance becomes \(2r\), the force becomes \(F/4\). If distance becomes \(3r\), the force becomes \(F/9\).

3. Force Between Multiple Charges

When more than two charges are present, each charge exerts a force on the selected charge. The total force is the vector sum of all these forces.

\[ \boxed{ \vec F_{\text{net}} = \vec F_1+\vec F_2+\vec F_3+\cdots+\vec F_n } \]

3.1 Superposition principle — step by step

Step 1: Select the charge on which force is required.

Step 2: Ignore all charges except one and calculate its force.

Step 3: Repeat this for every other charge.

Step 4: Give every force its correct direction.

Step 5: Add the forces as vectors.

For example, for a test charge \(q_0\) acted upon by charges \(q_1,q_2,q_3\):

\[ \boxed{ \vec F_0= \frac1{4\pi\varepsilon_0} \left[ \frac{q_0q_1}{r_1^2}\hat r_1+ \frac{q_0q_2}{r_2^2}\hat r_2+ \frac{q_0q_3}{r_3^2}\hat r_3 \right] } \]

4. Continuous Charge Distribution

Sometimes charge is spread continuously over a wire, surface or volume. Instead of considering every charge separately, we take a very small charge element \(dq\) and add all such elements using integration.

Distribution	Charge density	Small charge
Line	\(\lambda=Q/L\)	\(dq=\lambda\,dl\)
Surface	\(\sigma=Q/A\)	\(dq=\sigma\,dA\)
Volume	\(\rho=Q/V\)	\(dq=\rho\,dV\)

The field produced by a small element \(dq\) is

\[ d\vec E= \frac1{4\pi\varepsilon_0} \frac{dq}{r^2}\hat r. \]

Therefore the total field is

\[ \boxed{\vec E=\int d\vec E}. \]

Important: Before doing an integral, always decide the geometry, choose \(dq\), and use symmetry to find which components cancel.

5. Electric Field

The electric field at a point is the force experienced by a small positive test charge placed at that point divided by the test charge.

\[ \boxed{ \vec E=\lim_{q_0\to0}\frac{\vec F}{q_0} } \]

For most Class 12 calculations we write

\[ \boxed{\vec F=q_0\vec E}. \]

5.1 Electric field due to a point charge — complete derivation

Field at point \(P\) due to a positive point charge \(Q\).

Step 1: Let the source charge be \(Q\), the test charge be \(q_0\), and their separation be \(r\).

Step 2: Write Coulomb's force. \[ F= \frac1{4\pi\varepsilon_0} \frac{|Qq_0|}{r^2}. \]

Step 3: Definition of electric field. \[ E=\frac{F}{q_0}. \]

Step 4: Substitute \(F\). \[ E= \frac1{q_0} \frac1{4\pi\varepsilon_0} \frac{|Qq_0|}{r^2}. \]

Step 5: Cancel the test charge. \[ \boxed{ E= \frac1{4\pi\varepsilon_0} \frac{|Q|}{r^2} } \]

Vector form:

\[ \boxed{ \vec E= \frac1{4\pi\varepsilon_0} \frac{Q}{r^2}\hat r } \]

5.2 Direction

For \(Q>0\), the field is directed away from \(Q\). For \(Q<0\), the field is directed towards \(Q\).

6. Electric Field Lines

An electric field line is an imaginary line used to show the direction and relative strength of an electric field.

Field lines of a positive point charge.

Rules of electric field lines

1. They start from positive charge and end on negative charge or at infinity.
2. The tangent to a field line gives the direction of \(\vec E\).
3. Closer field lines represent a stronger field.
4. Electric field lines never cross each other.
5. In electrostatics, field lines do not form closed loops.
6. At the surface of a conductor in electrostatic equilibrium, field lines are perpendicular to the surface.

7. Electric Dipole

An electric dipole is a pair of equal and opposite charges separated by a small distance. Let the charges be \(-q\) and \(+q\), separated by \(2a\).

7.1 Dipole moment

Dipole moment is equal to the magnitude of either charge multiplied by the separation between the two charges.

\[ \boxed{p=q(2a)} \]

The direction of \(\vec p\) is from the negative charge to the positive charge. The SI unit is \(\mathrm{C\,m}\).

8. Electric Field Due to an Electric Dipole

8.1 Field on the axial line

The axial line is the straight line passing through both charges. Take a point \(P\) at distance \(r\) from the centre of the dipole.

Step 1: Distances.
Distance from \(+q\) to \(P\) is \(r-a\), while distance from \(-q\) to \(P\) is \(r+a\).

Step 2: Field due to \(+q\). \[ E_+= \frac1{4\pi\varepsilon_0} \frac{q}{(r-a)^2}. \]

Step 3: Field due to \(-q\). \[ E_-= \frac1{4\pi\varepsilon_0} \frac{q}{(r+a)^2}. \] At \(P\), this field is opposite to the field due to \(+q\).

Step 4: Net field. \[ E=E_+-E_-. \] Therefore, \[ E= \frac{q}{4\pi\varepsilon_0} \left[ \frac1{(r-a)^2} -\frac1{(r+a)^2} \right]. \]

Step 5: Take the LCM. \[ \frac1{(r-a)^2}-\frac1{(r+a)^2} = \frac{(r+a)^2-(r-a)^2} {(r-a)^2(r+a)^2}. \]

Step 6: Expand the numerator. \[ (r+a)^2-(r-a)^2 = (r^2+2ar+a^2)-(r^2-2ar+a^2) =4ar. \]

Step 7: Simplify the denominator. \[ (r-a)^2(r+a)^2=(r^2-a^2)^2. \] Thus, \[ E= \frac1{4\pi\varepsilon_0} \frac{4qar}{(r^2-a^2)^2}. \]

Step 8: Use dipole moment \(p=2aq\). Since \(4qa=2p\), \[ \boxed{ E_{\rm axial} = \frac1{4\pi\varepsilon_0} \frac{2pr}{(r^2-a^2)^2} }. \] This is the exact axial-field expression.

Step 9: Short dipole approximation.
If \(r\gg a\), then \(a\) is very small compared with \(r\), so \[ r^2-a^2\approx r^2. \] Hence \[ (r^2-a^2)^2\approx r^4. \] Therefore, \[ \boxed{ E_{\rm axial}\approx \frac1{4\pi\varepsilon_0} \frac{2p}{r^3} }. \]

8.2 Field on the equatorial line

The equatorial line is the perpendicular bisector of the dipole. Let \(P\) be at distance \(r\) from the centre.

Step 1: Distance from each charge. \[ R=\sqrt{r^2+a^2}. \]

Step 2: Field due to either charge. \[ E_0= \frac1{4\pi\varepsilon_0}\frac{q}{R^2}. \]

Step 3: Resolve the field.
From the right triangle, \[ \cos\alpha=\frac{a}{R}. \] The components perpendicular to the dipole axis cancel, while the components along the dipole axis add.

Step 4: Useful component of one field. \[ E_0\cos\alpha = \frac1{4\pi\varepsilon_0} \frac{q}{R^2}\frac{a}{R} = \frac1{4\pi\varepsilon_0} \frac{qa}{R^3}. \]

Step 5: Two components add. \[ E= \frac1{4\pi\varepsilon_0} \frac{2qa}{R^3}. \]

Step 6: Use \(p=2qa\) and \(R^3=(r^2+a^2)^{3/2}\). \[ \boxed{ E_{\rm equatorial} = \frac1{4\pi\varepsilon_0} \frac{p}{(r^2+a^2)^{3/2}} }. \]

Step 7: Short dipole approximation.
For \(r\gg a\), \[ (r^2+a^2)^{3/2}\approx r^3. \] Therefore, \[ \boxed{ E_{\rm equatorial} \approx \frac1{4\pi\varepsilon_0} \frac{p}{r^3} }. \] The direction is opposite to \(\vec p\).

Very important comparison for a short dipole: \[ \boxed{ E_{\rm axial}=2E_{\rm equatorial} } \] at the same distance \(r\) from the centre.

9. Torque on a Dipole in a Uniform Electric Field

Put an electric dipole in a uniform electric field \(\vec E\). Let the dipole moment \(\vec p\) make an angle \(\theta\) with the field.

Step 1: Force on \(+q\). \[ \vec F_+=q\vec E. \]

Step 2: Force on \(-q\). \[ \vec F_-=-q\vec E. \]

Step 3: Net force. \[ \vec F_{\rm net} = q\vec E-q\vec E =\boxed{0}. \] So a dipole in a uniform electric field has zero net force.

Step 4: But torque is present.
The two forces act along different parallel lines. They form a couple and rotate the dipole.

Step 5: Perpendicular separation. The perpendicular distance between the two lines of action is \[ 2a\sin\theta. \]

Step 6: Torque of the couple. \[ \tau = qE(2a\sin\theta). \]

Step 7: Use \(p=2aq\). \[ \boxed{\tau=pE\sin\theta}. \]

Step 8: Vector form. \[ \boxed{\vec\tau=\vec p\times\vec E}. \]

9.1 Special cases

\(\theta\)	\(\tau=pE\sin\theta\)	Meaning
\(0^\circ\)	0	Dipole is parallel to field.
\(90^\circ\)	\(pE\)	Maximum torque.
\(180^\circ\)	0	Dipole is opposite to field.

9.2 Potential energy of a dipole

\[ \boxed{U=-\vec p\cdot\vec E=-pE\cos\theta} \]

10. Electric Flux

Electric flux tells us how much electric field passes through a surface. For a uniform electric field crossing a flat surface of area \(A\):

\[ \boxed{\Phi_E=EA\cos\theta} \]

Here \(\theta\) is the angle between the electric field \(\vec E\) and the normal to the surface.

10.1 Why \(EA\cos\theta\)?

Step 1: Only the component of the field perpendicular to the surface passes through the surface.

Step 2: Perpendicular component is \[ E_\perp=E\cos\theta. \]

Step 3: Multiply this component by area \(A\): \[ \Phi_E=E_\perp A. \]

Step 4: \[ \boxed{\Phi_E=EA\cos\theta}. \]

10.2 Area vector

For a flat surface, the area vector is \[ \vec A=A\hat n, \] where \(\hat n\) is the unit normal. Therefore,

\[ \boxed{\Phi_E=\vec E\cdot\vec A}. \]

10.3 Closed surface

For a closed surface:

\[ \boxed{ \Phi_E=\oint\vec E\cdot d\vec A } \]

11. Gauss's Theorem

Gauss's theorem, also called Gauss's law, gives the total electric flux through a closed surface in terms of the charge enclosed by that surface.

\[ \boxed{ \oint\vec E\cdot d\vec A = \frac{Q_{\rm enclosed}}{\varepsilon_0} } \]

11.1 Meaning in simple words

Take any closed imaginary surface. Add the electric flux through every small part of that surface. The result depends only on the total charge inside the surface.

11.2 Step-by-step use of Gauss's law

Step 1: Look at the symmetry of the charge distribution.

Step 2: Select a suitable closed Gaussian surface.

Step 3: Find the direction of electric field.

Step 4: Calculate \(\vec E\cdot d\vec A\) on each part of the surface.

Step 5: Find the total enclosed charge.

Step 6: Apply \[ \oint\vec E\cdot d\vec A=\frac{Q_{\rm enc}}{\varepsilon_0}. \]

Step 7: Solve for \(E\).

Very important: Gauss's law is true for every closed surface. But it is especially useful for calculating \(E\) when the charge distribution has spherical, cylindrical or planar symmetry.

12. Field Due to an Infinitely Long Straight Charged Wire

Consider an infinitely long straight wire with uniform linear charge density \(\lambda\). Find the electric field at a distance \(r\) from the wire.

Step 1: Choose the Gaussian surface.
Take a cylinder of radius \(r\) and length \(L\), with the wire along its axis.

Step 2: Use symmetry.
Because the wire is infinitely long and uniformly charged, \(E\) has the same magnitude at every point on the curved surface at the same distance \(r\).

Step 3: Flux through curved surface.
On the curved surface, \(\vec E\) and \(d\vec A\) are parallel. Therefore, \[ \vec E\cdot d\vec A=E\,dA. \] The curved area is \[ A_{\rm curved}=2\pi rL. \] Hence, \[ \Phi_{\rm curved}=E(2\pi rL). \]

Step 4: Flux through the two end caps.
At the end caps, the electric field is parallel to the plane of the cap. Therefore it is perpendicular to the area vector. \[ \vec E\cdot d\vec A=0. \] So the end caps contribute zero flux.

Step 5: Total flux. \[ \Phi=E(2\pi rL). \]

Step 6: Enclosed charge.
Linear charge density is \[ \lambda=\frac{Q}{L}. \] Therefore, \[ Q_{\rm enc}=\lambda L. \]

Step 7: Apply Gauss's law. \[ E(2\pi rL) = \frac{\lambda L}{\varepsilon_0}. \]

Step 8: Cancel \(L\). \[ 2\pi rE=\frac{\lambda}{\varepsilon_0}. \]

Step 9: Final answer. \[ \boxed{ E=\frac{\lambda}{2\pi\varepsilon_0r} } \]

For an infinite line charge: \[ \boxed{E\propto\frac1r}. \] The field is radially outward for positive \(\lambda\) and radially inward for negative \(\lambda\).

13. Field Due to a Uniformly Charged Infinite Plane Sheet

Let an infinite plane sheet have uniform surface charge density \(\sigma\). We want the electric field near the sheet.

Step 1: Use symmetry.
For an infinite uniformly charged sheet, the field must be perpendicular to the sheet. The magnitude is the same at equal distances on both sides.

Step 2: Choose a Gaussian surface.
Take a small cylindrical pillbox of cross-sectional area \(A\), with the sheet passing through its middle.

Step 3: Flux through the curved surface.
The field is parallel to the curved surface, so \[ \vec E\cdot d\vec A=0. \] Thus the curved surface gives zero flux.

Step 4: Flux through the two flat faces.
On each flat face, field and area vector are parallel: \[ \Phi_1=EA,\qquad \Phi_2=EA. \] Therefore, \[ \Phi=2EA. \]

Step 5: Enclosed charge.
Surface charge density is \[ \sigma=\frac{Q}{A}. \] Hence, \[ Q_{\rm enc}=\sigma A. \]

Step 6: Apply Gauss's law. \[ 2EA=\frac{\sigma A}{\varepsilon_0}. \]

Step 7: Cancel \(A\). \[ 2E=\frac{\sigma}{\varepsilon_0}. \]

Step 8: Final answer. \[ \boxed{ E=\frac{\sigma}{2\varepsilon_0} } \]

Special result: For an ideal infinite plane sheet, \[ \boxed{E=\frac{\sigma}{2\varepsilon_0}} \] and the field is independent of distance from the sheet.

14. Field Due to a Uniformly Charged Thin Spherical Shell

Consider a thin spherical shell of radius \(R\) carrying total charge \(Q\), uniformly distributed over its surface. We find the field in two regions: inside and outside.

14.1 Outside the shell: \(r>R\)

Step 1: Choose Gaussian surface.
Take a spherical Gaussian surface of radius \(r\), concentric with the charged shell.

Step 2: Use spherical symmetry.
The electric field has the same magnitude \(E\) at every point of the Gaussian sphere and is radial.

Step 3: Calculate surface area. \[ A=4\pi r^2. \]

Step 4: Calculate flux.
Here \(\vec E\) is parallel to \(d\vec A\), so \[ \Phi=E(4\pi r^2). \]

Step 5: Enclosed charge. \[ Q_{\rm enc}=Q. \]

Step 6: Apply Gauss's law. \[ E(4\pi r^2)=\frac{Q}{\varepsilon_0}. \]

Step 7: Solve for \(E\). \[ \boxed{ E= \frac1{4\pi\varepsilon_0} \frac{Q}{r^2} } \qquad (r>R). \]

Therefore, outside the shell, the shell behaves like a point charge \(Q\) placed at its centre.

14.2 Inside the shell: \(r<r\)< h3="">

Step 1: Choose Gaussian surface.
Take a concentric spherical Gaussian surface of radius \(r<r\). <="" div="">

Step 2: Find enclosed charge.
All charge is on the shell at radius \(R\). The Gaussian sphere lies inside it. Therefore, \[ \boxed{Q_{\rm enc}=0}. \]

Step 3: Apply Gauss's law. \[ \oint\vec E\cdot d\vec A = \frac{0}{\varepsilon_0}=0. \]

Step 4: Use spherical symmetry. \[ E(4\pi r^2)=0. \] Since \(4\pi r^2\neq0\), \[ \boxed{E=0}. \]

14.3 Field just outside the shell

At \(r=R^+\):

\[ \boxed{ E(R^+)= \frac1{4\pi\varepsilon_0} \frac{Q}{R^2} } \]

14.4 Complete result

Region	Electric field
\(r<r\)< td="">	\(\boxed{E=0}\)
\(r=R^+\)	\(\boxed{E=\dfrac{Q}{4\pi\varepsilon_0R^2}}\)
\(r>R\)	\(\boxed{E=\dfrac{Q}{4\pi\varepsilon_0r^2}}\)

15. Complete Formula Revision

Concept	Formula
Quantisation of charge	\(q=ne\)
Coulomb's law	\(F=\dfrac1{4\pi\varepsilon_0}\dfrac{|q_1q_2|}{r^2}\)
Superposition	\(\vec F_{\rm net}=\sum_i\vec F_i\)
Electric field	\(\vec E=\vec F/q_0\)
Point charge field	\(\vec E=\dfrac1{4\pi\varepsilon_0}\dfrac{Q}{r^2}\hat r\)
Force in electric field	\(\vec F=q\vec E\)
Dipole moment	\(p=2aq\)
Axial field, short dipole	\(E_{\rm axial}=\dfrac1{4\pi\varepsilon_0}\dfrac{2p}{r^3}\)
Equatorial field, short dipole	\(E_{\rm equatorial}=\dfrac1{4\pi\varepsilon_0}\dfrac{p}{r^3}\)
Dipole torque	\(\vec\tau=\vec p\times\vec E\)
Torque magnitude	\(\tau=pE\sin\theta\)
Dipole potential energy	\(U=-pE\cos\theta\)
Electric flux	\(\Phi_E=\vec E\cdot\vec A=EA\cos\theta\)
Gauss's law	\(\oint\vec E\cdot d\vec A=\dfrac{Q_{\rm enc}}{\varepsilon_0}\)
Infinite line charge	\(E=\dfrac{\lambda}{2\pi\varepsilon_0r}\)
Infinite plane sheet	\(E=\dfrac{\sigma}{2\varepsilon_0}\)
Spherical shell inside	\(E=0\)
Spherical shell outside	\(E=\dfrac1{4\pi\varepsilon_0}\dfrac{Q}{r^2}\)

Most important derivations to practise

1. Electric field due to a point charge.
2. Electric field due to an electric dipole on the axial line.
3. Electric field due to an electric dipole on the equatorial line.
4. Torque on an electric dipole in a uniform electric field.
5. Electric flux and Gauss's law.
6. Electric field due to an infinitely long straight wire.
7. Electric field due to an infinite plane sheet.
8. Electric field inside and outside a uniformly charged thin spherical shell.

Quick memory map:
Point charge → \(1/r^2\)
Infinite line charge → \(1/r\)
Infinite plane sheet → constant \(E\)
Short dipole → \(1/r^3\)
Spherical shell inside → \(E=0\)

Common mistakes

• Do not forget that force is a vector.
• In flux, \(\theta\) is measured from the surface normal.
• Dipole moment points from \(-q\) to \(+q\).
• Do not use the short-dipole formula unless \(r\gg a\).
• For Gauss's law, use the net enclosed charge.
• For an infinite sheet, field does not decrease with distance.
• For an infinite line charge, field varies as \(1/r\).
• Inside a uniformly charged thin spherical shell, \(E=0\).

SEO topic coverage: Class 12 Physics Electric Charges and Fields, Electric Charges and Fields notes, Coulomb's law, electric field, electric field lines, electric dipole, dipole moment, electric dipole field, torque on dipole, electric flux, Gauss's law, infinite line charge, infinite plane sheet and uniformly charged spherical shell.

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Chapter 2 — Electrostatic Potential and Capacitance

Potential gives an energy-based way to describe electrostatic interactions. This chapter connects electric field with potential, develops equipotential surfaces and potential energy, then builds the capacitor model and its energy storage.

Chapter 2 — Electrostatic Potential and Capacitance

This chapter explains electric potential, potential difference, potential due to different charge arrangements, equipotential surfaces, potential energy, conductors, dielectrics, capacitors and capacitance. The language is kept very simple and the important formulas are highlighted for quick revision.

Chapter in one line: Electric field tells us about force; electric potential tells us about work done per unit charge; a capacitor stores charge and electrical energy.

Exam focus: Learn definitions, units, sign conventions, standard formulas, series/parallel capacitor relations, parallel-plate capacitor formula and short conceptual differences. For this scope, energy stored in a capacitor is formula-based; no derivation is included.

Chapter contents

1. Electric Potential and Potential Difference
2. Potential Due to a Point Charge
3. Potential Due to an Electric Dipole
4. Potential Due to a System of Charges
5. Equipotential Surfaces
6. Potential Energy
7. Electrostatics of Conductors
8. Dielectrics and Polarisation
9. Capacitors and Capacitance
10. Parallel Plate Capacitor
11. Combination of Capacitors
12. Energy Stored in a Capacitor — Formulae Only
13. Complete Formula Revision

1. Electric Potential and Potential Difference

Definition: Electric potential at a point is the work done per unit positive test charge in bringing it from infinity to that point, without acceleration.

Simple idea: Electric field tells us the force per unit charge. Electric potential tells us the work or energy per unit charge.

\[ \boxed{V=\frac{W}{q}} \]

where \(W\) is the work done and \(q\) is the test charge.

1.1 SI unit

The SI unit of electric potential is volt (V).

\[ \boxed{1\ \mathrm{V}=1\ \mathrm{J\,C^{-1}}} \]

1.2 Potential difference

The potential difference between two points is the work done per unit positive test charge in moving the charge from one point to another.

\[ \boxed{V_B-V_A=\frac{W_{A\to B}}{q}} \]

More generally, the work done by the electrostatic field is

\[ \boxed{W_{\text{field}}=q(V_A-V_B)} \]

Remember: Potential is a scalar quantity. It has magnitude and sign, but no direction.

1.3 Relation between electric field and potential

In a one-dimensional situation, electric field is the negative rate of change of potential with distance.

\[ \boxed{E=-\frac{dV}{dr}} \]

The negative sign means the potential decreases in the direction of the electric field.

Easy memory: Field points from higher potential to lower potential.

2. Electric Potential Due to a Point Charge

Consider a point charge \(q\). Let a point \(P\) be at distance \(r\) from the charge.

\[ \boxed{V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}} \]

Using \(k=\dfrac{1}{4\pi\varepsilon_0}\):

\[ \boxed{V=\frac{kq}{r}} \]

2.1 Important points

• Potential is positive for a positive source charge.
• Potential is negative for a negative source charge.
• Potential becomes zero at infinity by the usual reference convention.
• Potential varies as \(1/r\).

2.2 Short derivation

Step 1: Electric field due to point charge \(q\): \[ E=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}. \]

Step 2: Potential difference from infinity to \(r\): \[ V(r)-V(\infty)=-\int_{\infty}^{r}\vec E\cdot d\vec r. \]

Step 3: Taking \(V(\infty)=0\): \[ V(r)=\frac{q}{4\pi\varepsilon_0}\int_r^\infty\frac{dr}{r^2}. \]

Step 4: \[ \boxed{V(r)=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}}. \]

3. Electric Potential Due to an Electric Dipole

Electric dipole: A pair of equal and opposite charges separated by a small distance.

If the charges are \(+q\) and \(-q\), separated by distance \(2a\), the dipole moment is

\[ \boxed{\vec p=q(2a)\,\hat p} \]

The direction of \(\vec p\) is from negative charge to positive charge.

3.1 Potential at a general point

The potential due to a dipole is the algebraic sum of the potentials due to its two charges. For a distant point making angle \(\theta\) with the dipole axis:

\[ \boxed{ V=\frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2} } \qquad (r\gg a) \]

3.2 On the axial line

On the axial line, \(\theta=0^\circ\) on the positive side, so \(\cos\theta=1\).

\[ \boxed{V_{\text{axial}}\approx \frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}} \]

3.3 On the equatorial line

On the equatorial line, \(\theta=90^\circ\), so \(\cos\theta=0\).

\[ \boxed{V_{\text{equatorial}}=0} \]

Very important: Dipole potential on the equatorial line is zero, while electric field there is not zero.

4. Potential Due to a System of Charges

Because electric potential is a scalar quantity, the total potential is the ordinary algebraic sum of the potentials due to all source charges.

\[ \boxed{ V=\frac{1}{4\pi\varepsilon_0} \left( \frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}+\cdots \right) } \]

For \(n\) point charges:

\[ \boxed{ V=\frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\frac{q_i}{r_i} } \]

Why is this easy? Force is a vector, so directions must be added. Potential is a scalar, so only signs and magnitudes are added.

5. Equipotential Surfaces

Definition: An equipotential surface is a surface on which electric potential is the same at every point.

\[ \boxed{V=\text{constant}} \]

5.1 Properties

1. No work is done in moving a test charge along an equipotential surface.
2. Electric field is perpendicular to an equipotential surface.
3. Two equipotential surfaces never intersect.
4. Where equipotential surfaces are closer, the electric field is stronger.

5.2 Why no work is done?

\[ W=q(V_A-V_B) \]

On the same equipotential surface, \(V_A=V_B\). Therefore:

\[ \boxed{W=0} \]

Memory trick: Equipotential → Equal V → Zero work along the surface → Field at right angle.

6. Electrical Potential Energy

6.1 Two point charges

The electrical potential energy of a system of two point charges \(q_1\) and \(q_2\), separated by distance \(r\), is

\[ \boxed{ U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r} } \]

or

\[ \boxed{U=q_2V_1} \]

where \(V_1\) is the potential at the position of \(q_2\) due to \(q_1\).

6.2 System of many point charges

\[ \boxed{ U=\frac{1}{4\pi\varepsilon_0} \sum_{i<j}\frac{q_iq_j}{r_{ij}} \]<="" div="" }="">

6.3 Dipole in an external electric field

For an electric dipole of moment \(p\) in a uniform external field \(E\):

\[ \boxed{U=-pE\cos\theta} \]

Therefore:

• At \(\theta=0^\circ\): \(U=-pE\), minimum.
• At \(\theta=90^\circ\): \(U=0\).
• At \(\theta=180^\circ\): \(U=+pE\), maximum.

Memory: U = −pE cosθ. Stable equilibrium is when the dipole is parallel to the field.

7. Electrostatics of Conductors

A conductor contains charges that can move freely. In electrostatic equilibrium, these charges arrange themselves so that the conductor reaches a state with no further movement of free charge.

7.1 Electric field inside a conductor

In electrostatic equilibrium: \[ \boxed{E_{\text{inside conductor}}=0} \]

7.2 Why?

If a non-zero electric field existed inside the conductor, free charges would experience force and move. That would contradict electrostatic equilibrium. Hence the electric field inside is zero.

7.3 Potential of a conductor

Since \(E=0\) inside a conductor, the potential is constant throughout the conductor.

\[ \boxed{V=\text{constant inside a conductor}} \]

7.4 Charge on a conductor

• Excess charge resides on the surface of a conductor in electrostatic equilibrium.
• The electric field just outside a charged conductor is normal to its surface.
• The electric field has no tangential component at the surface in electrostatic equilibrium.

7.5 Free and bound charges

Free charges can move through a conductor. In an ideal electrostatic conductor, they rearrange until equilibrium is reached. In an insulator or dielectric, charges are largely bound to atoms or molecules and cannot move freely through the material.

Must remember: Inside a conductor in electrostatic equilibrium: E = 0 and V = constant.

8. Dielectrics and Polarisation

Dielectric: A dielectric is an insulating material that does not allow free charge to move easily through it, but its molecules can become polarised in an electric field.

8.1 Polarisation

When a dielectric is placed in an external electric field, positive and negative bound charges shift slightly in opposite directions. This produces induced dipoles. The process is called polarisation.

\[ \boxed{\vec P=\frac{\text{dipole moment}}{\text{volume}}} \]

Here \(\vec P\) is the polarisation vector.

8.2 Effect of dielectric

A dielectric placed between capacitor plates reduces the effective electric field inside the material compared with the vacuum case, and it increases the capacitance of the capacitor.

\[ \boxed{C=K C_0} \]

where \(K\) is the dielectric constant (relative permittivity) and \(C_0\) is the capacitance without dielectric.

9. Capacitors and Capacitance

Capacitor: A capacitor is a system of two conductors separated by an insulating material or empty space, used to store electric charge and electrical energy.

9.1 Capacitance

Capacitance is defined as the charge stored per unit potential difference.

\[ \boxed{C=\frac{Q}{V}} \]

SI unit: farad (F).

\[ \boxed{1\ \mathrm F=1\ \mathrm{C\,V^{-1}}} \]

Important: Capacitance depends on the geometry and medium of the capacitor. For an ideal capacitor, \(C\) does not depend on the amount of charge stored.

10. Parallel Plate Capacitor

A parallel plate capacitor consists of two large conducting plates of area \(A\), separated by a small distance \(d\). For an ideal capacitor, edge effects are neglected.

10.1 Without dielectric

\[ \boxed{ C_0=\frac{\varepsilon_0 A}{d} } \]

10.2 With dielectric completely filling the space

\[ \boxed{ C=K\frac{\varepsilon_0 A}{d} } \]

Since \(K=\varepsilon_r\):

\[ \boxed{C=\frac{\varepsilon A}{d}} \]

where \(\varepsilon=K\varepsilon_0\).

10.3 What changes when dielectric is inserted?

Quantity	Without dielectric	With dielectric \(K\)
Capacitance	\(C_0\)	\(KC_0\)
Electric field	\(E_0\)	Reduced for the isolated charge configuration
Permittivity	\(\varepsilon_0\)	\(\varepsilon=K\varepsilon_0\)

Key formula: For a parallel plate capacitor, capacitance increases with plate area and decreases with plate separation.

11. Combination of Capacitors

11.1 Capacitors in parallel

In parallel, all capacitors have the same potential difference.

\[ \boxed{V=V_1=V_2=V_3=\cdots} \]

The total charge is the sum:

\[ Q=Q_1+Q_2+Q_3+\cdots \]

Therefore:

\[ \boxed{C_{\text{eq}}=C_1+C_2+C_3+\cdots} \]

Parallel memory: Capacitances add directly.

11.2 Capacitors in series

In series, the magnitude of charge on each capacitor is the same:

\[ \boxed{Q=Q_1=Q_2=Q_3=\cdots} \]

Total potential difference is:

\[ V=V_1+V_2+V_3+\cdots \]

Therefore:

\[ \boxed{ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\cdots } \]

For two capacitors:

\[ \boxed{ C_{\text{eq}}=\frac{C_1C_2}{C_1+C_2} } \]

Series memory: Reciprocals add.

11.3 Series vs parallel

Feature	Series	Parallel
Charge	Same on each capacitor	Charge divides
Potential difference	Divides	Same across each
Equivalent capacitance	Reciprocals add	Direct addition
For identical capacitors \(C\), \(n\) in combination	\(C_{\rm eq}=C/n\)	\(C_{\rm eq}=nC\)

12. Energy Stored in a Capacitor — Formulae Only

As required for this syllabus scope, remember the standard formulae. No derivation is needed here.

Basic form

\[ \boxed{U=\frac12QV} \]

Using capacitance

\[ \boxed{U=\frac12CV^2} \]

Using charge

\[ \boxed{U=\frac{Q^2}{2C}} \]

Energy density

\[ \boxed{u=\frac12\varepsilon E^2} \]

13. Complete Formula Revision

Topic	Formula
Potential	\(V=W/q\)
Potential difference	\(V_B-V_A=W_{A\to B}/q\)
Point charge	\(V=kq/r\)
Dipole potential	\(V=k p\cos\theta/r^2\) for \(r\gg a\)
System of charges	\(V=k\sum q_i/r_i\)
Field-potential relation	\(E=-dV/dr\)
Two-charge potential energy	\(U=kq_1q_2/r\)
Dipole energy	\(U=-pE\cos\theta\)
Capacitance	\(C=Q/V\)
Parallel plate capacitor	\(C_0=\varepsilon_0A/d\)
With dielectric	\(C=K\varepsilon_0A/d\)
Parallel combination	\(C_{\rm eq}=C_1+C_2+\cdots\)
Series combination	\(1/C_{\rm eq}=1/C_1+1/C_2+\cdots\)
Two capacitors in series	\(C_{\rm eq}=C_1C_2/(C_1+C_2)\)
Energy	\(U=\frac12QV=\frac12CV^2=\frac{Q^2}{2C}\)
Energy density	\(u=\frac12\varepsilon E^2\)

One-minute revision:
Potential = work per unit charge.
Point charge potential = \(kq/r\).
Dipole potential = \(kp\cos\theta/r^2\) (far point).
Equipotential surface = same potential; work along it = zero.
Conductor in electrostatic equilibrium = \(E=0\), \(V=\) constant inside.
Capacitance = \(Q/V\).
Parallel capacitors = capacitances add.
Series capacitors = reciprocals add.
Parallel plate capacitor = \(\varepsilon A/d\).
Capacitor energy = \(\frac12CV^2\).

Exam tip: In numerical questions, first write the given quantities, convert SI units, write the relevant formula, substitute carefully, and finally write the unit.

Source alignment: The chapter organisation follows the NCERT Class 12 Physics “Electrostatic Potential and Capacitance” chapter and the electrostatics syllabus topics: potential, dipole, equipotential surfaces, potential energy, conductors, dielectrics, capacitors, combinations, parallel-plate capacitor and energy stored.

High-Yield Derivation 1: Relation Between Electric Field and Potential

Given: A small positive test charge moves through a small distance dr in an electric field.

Step 1: Work done by the electric field over a small displacement is
\[dW_{\text{field}}=qE\,dr\]

Step 2: Potential change is defined from the work done by an external agent as
\[dV=-\frac{dW_{\text{field}}}{q}\]

Step 3: Substitute \(dW_{\text{field}}=qE\,dr\):
\[dV=-E\,dr\]

Step 4: Therefore,
\[\boxed{E=-\frac{dV}{dr}}\]

Meaning: Electric field points in the direction of decreasing potential. A steeper potential change means a stronger electric field.

High-Yield Derivation 2: Potential of an Electric Dipole at a Far Point

Given: A dipole has charges \(+q\) and \(-q\), separation \(2a\), dipole moment \(p=2aq\). A point \(P\) is at distance \(r\), where \(r\gg a\).

Step 1: Potential at \(P\) is the algebraic sum of potentials due to the two charges:
\[V=k\left(\frac{q}{r_+}-\frac{q}{r_-}\right)\]

Step 2: For a far point, use the standard dipole approximation:
\[\frac{1}{r_+}-\frac{1}{r_-}\approx\frac{2a\cos\theta}{r^2}\]

Step 3: Hence,
\[V=k\frac{2aq\cos\theta}{r^2}\]

Step 4: Since \(p=2aq\),
\[\boxed{V=\frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}}\]

Important: This far-field expression is used when \(r\gg a\).

High-Yield Derivation 3: Capacitance of a Parallel-Plate Capacitor

Given: Two large parallel plates of area \(A\), separation \(d\), charges \(+Q\) and \(-Q\), with vacuum between them.

Step 1: Surface charge density is
\[\sigma=\frac{Q}{A}\]

Step 2: The electric field between large oppositely charged plates is approximately
\[E=\frac{\sigma}{\varepsilon_0}=\frac{Q}{\varepsilon_0A}\]

Step 3: Potential difference between the plates is
\[V=Ed=\frac{Qd}{\varepsilon_0A}\]

Step 4: Capacitance is \(C=Q/V\). Therefore,
\[\boxed{C=\frac{\varepsilon_0A}{d}}\]

Conclusion: Larger plate area increases capacitance; larger separation decreases it.

High-Yield Derivation 4: Energy Stored in a Capacitor

Idea: Charging a capacitor requires work because each additional small charge is brought to an increasing potential.

Step 1: At an intermediate charge \(q\), potential is \(V=q/C\).

Step 2: Small work needed to add \(dq\) is
\[dU=V\,dq=\frac{q}{C}dq\]

Step 3: Integrate from \(q=0\) to \(q=Q\):
\[U=\int_0^Q\frac{q}{C}dq=\frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q\]

Step 4: Therefore,
\[\boxed{U=\frac{Q^2}{2C}}\]

Step 5: Since \(Q=CV\), equivalent forms are
\[\boxed{U=\frac12CV^2=\frac12QV}\]

Unit I Solved Numerical Problems

Use this standard board method: Given → Required → Formula → Substitution → Calculation → Final answer with unit.

Example 1 — Potential of a Point Charge

Question: Find the potential at a point 0.20 m from a charge of \(+4\,\mu C\).

Given: \(q=4\times10^{-6}C\), \(r=0.20m\).

Formula: \(V=kq/r\), where \(k=9\times10^9\,Nm^2/C^2\).

Substitution: \[V=\frac{(9\times10^9)(4\times10^{-6})}{0.20}=1.8\times10^5V\]

Answer: \(\boxed{1.8\times10^5\,V}\)

Example 2 — Equivalent Capacitance in Parallel

Question: Capacitors of \(2\,\mu F\) and \(3\,\mu F\) are connected in parallel. Find equivalent capacitance.

Formula: \(C_p=C_1+C_2\).

\[C_p=2+3=5\,\mu F\]

Answer: \(\boxed{5\,\mu F}\)

Example 3 — Energy Stored in a Capacitor

Question: A \(10\,\mu F\) capacitor is charged to \(100\,V\). Find stored energy.

Formula: \(U=\frac12CV^2\).

\[U=\frac12(10\times10^{-6})(100)^2=0.05J\]

Answer: \(\boxed{5.0\times10^{-2}\,J}\)

Unit I Master Revision — Electrostatics

Must-Know Derivations

• Electric field due to a point charge.
• Electric dipole field on axial and equatorial positions.
• Torque on a dipole in a uniform electric field.
• Gauss-law applications: line charge, plane sheet, spherical shell.
• Potential due to a point charge and dipole.
• Relation \(E=-dV/dr\).
• Capacitance of a parallel-plate capacitor.
• Energy stored in a capacitor.

Common Exam Mistakes

• Confusing electric field (vector) with electric potential (scalar).
• Forgetting the sign of potential due to a negative charge.
• Using dipole far-field formulas when \(r\) is not much larger than dipole size.
• Mixing series and parallel capacitor rules.
• Forgetting that an equipotential surface is perpendicular to the electric field.
• Using Gauss's law without checking symmetry.
• Dropping units or powers of ten in numerical answers.

One-Page Formula Sheet

Coulomb
\(F=k\frac{|q_1q_2|}{r^2}\)

Field
\(E=k\frac{|q|}{r^2}\)

Flux
\(\Phi_E=EA\cos\theta\)

Gauss
\(\oint\vec E\cdot d\vec A=Q_{enc}/\varepsilon_0\)

Potential
\(V=kq/r\)

Dipole potential
\(V=kp\cos\theta/r^2\)

Energy
\(U=kq_1q_2/r\)

Capacitance
\(C=Q/V\)

Parallel plate
\(C=\varepsilon_0A/d\)

Capacitor energy
\(U=\frac12CV^2\)

Dielectric
\(C=K\varepsilon_0A/d\)

Energy density
\(u=\frac12\varepsilon E^2\)

Board Answer-Writing Pattern

1. Write the relevant law/definition first.
2. Draw a clean labelled diagram when geometry is involved.
3. Write every important algebraic step in a derivation.
4. Keep vectors, signs and directions clear.
5. Box the final result and include SI units in numericals.

Class 12 Physics Electrostatics — Frequently Asked Questions

What is included in Class 12 Physics Unit 1 Electrostatics?

Unit 1 covers Electric Charges and Fields and Electrostatic Potential and Capacitance. These notes cover the concepts, important formulas, derivations, diagrams and numerical methods needed for systematic board preparation.

Which derivations should I practise from Electrostatics?

Give special attention to Coulomb-law applications, electric-field results, electric dipole fields, torque on a dipole, Gauss-law applications, potential due to charges and dipoles, parallel-plate capacitor capacitance and capacitor energy.

Are these Class 12 Physics notes mobile friendly?

Yes. The page uses a responsive layout, readable typography and responsive MathJax equations so the notes can be studied comfortably on phones and tablets.

Where can I find more free study material?

Visit UniversityScope.com for free study resources, previous-year papers and exam-preparation material.

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Class 12 Physics Unit I — Electrostatics • Board-focused concept mastery, derivations, formulas and numerical practice.