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Electrostatic Potential and Capacitance

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Chapter Notes

Overview
This chapter builds on the concept of conservative forces from Class XI to introduce electrostatic potential and potential energy. The central idea is that work done against the electrostatic force in moving a charge is stored as potential energy, and the potential at a point is the work done per unit charge in bringing a test charge from infinity. Students must understand that potential is a scalar quantity, potential difference is path-independent, and that conductors in electrostatics have zero internal field and constant potential. This chapter forms the foundation for understanding capacitors, energy storage, and electric circuits.
Key Concepts (12)
Electrostatic Potential Energy
The work done by an external force in bringing a charge q from infinity to a point in an electrostatic field, without acceleration, is stored as the electrostatic potential energy of the charge at that point.
e.g. Bringing a positive test charge from far away towards a fixed positive charge requires work against repulsion; this work becomes potential energy.
Electrostatic Potential (V)
The work done by an external force in bringing a unit positive test charge from infinity to a point in an electrostatic field, without acceleration.
e.g. If 4 J of work is done to bring a +1 C charge from infinity to a point, the potential at that point is 4 V.
Equipotential Surface
A surface over which the electrostatic potential is constant at every point. No work is done in moving a charge on an equipotential surface.
e.g. For a point charge, concentric spherical surfaces centered at the charge are equipotential surfaces.
Dipole Moment (p)
A vector quantity defined as the product of the magnitude of either charge and the separation between them, directed from the negative to the positive charge. p = q × 2a.
e.g. A water molecule has a permanent dipole moment due to its bent shape.
Potential Energy of a System of Charges
The total work done in assembling the charges from infinity to their given positions, stored as electrostatic potential energy of the system.
e.g. For two like charges, potential energy is positive because work is done against repulsion.
Conservative Force
A force for which the work done in moving a particle between two points is independent of the path taken. Coulomb force and gravitational force are conservative.
e.g. The work done by the electrostatic field in moving a charge from R to P depends only on the initial and final positions, not on the path.
Electrostatic Field Inside a Conductor
In electrostatic equilibrium, the electric field inside a conductor is zero everywhere. Free charges redistribute to cancel any internal field.
e.g. A charged metal sphere has zero electric field inside it.
Excess Charge on a Conductor
In electrostatic equilibrium, any excess charge on a conductor resides entirely on its surface, not in the interior.
e.g. If you charge a hollow metal sphere, all charge lies on its outer surface.
Potential Due to a Point Charge
The electrostatic potential at a distance r from a point charge Q is V = (1/4πε0) (Q/r).
e.g. For Q = 4 × 10⁻⁷ C at r = 0.09 m, V = 4 × 10⁴ V.
Potential Due to an Electric Dipole
At a point far from the dipole (r >> a), the potential is V = (1/4πε0) (p cosθ / r²) = (1/4πε0) (p·r̂ / r²).
e.g. On the dipole axis (θ = 0), V = (1/4πε0) (p/r²); in the equatorial plane (θ = 90°), V = 0.
Potential Energy of a Dipole in an External Field
The potential energy of a dipole with moment p in a uniform external field E is U = –p·E = –pE cosθ, where θ is the angle between p and E.
e.g. When the dipole is aligned with the field (θ = 0°), potential energy is minimum (–pE).
Electron Volt (eV)
The energy gained by an electron when accelerated through a potential difference of 1 volt. 1 eV = 1.6 × 10⁻¹⁹ J.
e.g. An electron accelerated through 100 V gains 100 eV of kinetic energy.
Formulas & Equations (14)
W_RP = ∫_R^P F_ext · dl = –∫_R^P F_e · dl
Reads as: Work done by external force from R to P equals the negative of work done by electric force.
Variables: W_RP: work done by external force (J); F_ext: external force (N); F_e: electric force (N); dl: infinitesimal displacement (m).
Use when: Defining potential energy difference between two points.
⚠ Students forget the negative sign and think work done by external force equals work done by electric field.
ΔU = U_P – U_R = W_RP
Reads as: Change in potential energy equals work done by external force.
Variables: ΔU: potential energy difference (J); U_P: potential energy at P (J); U_R: potential energy at R (J); W_RP: work done by external force (J).
Use when: Calculating potential energy difference from work done.
⚠ Students write ΔU = –W_RP, confusing with work done by electric field.
V_P – V_R = (U_P – U_R)/q = –∫_R^P E · dl
Reads as: Potential difference between P and R equals work done per unit charge by external force.
Variables: V_P: potential at P (V); V_R: potential at R (V); q: test charge (C); E: electric field (N/C or V/m); dl: displacement (m).
Use when: Defining potential difference from electric field.
⚠ Students forget the negative sign in the integral form.
V(r) = (1/4πε₀) (Q/r)
Reads as: Potential at distance r from a point charge Q is one over four pi epsilon naught times Q over r.
Variables: V: potential (V); ε₀: permittivity of free space (8.854 × 10⁻¹² C²/N·m²); Q: source charge (C); r: distance from charge (m).
Use when: Finding potential due to a single point charge.
⚠ Students forget that r is the distance from the charge, not the coordinate.
V = (1/4πε₀) (p cosθ / r²) = (1/4πε₀) (p·r̂ / r²)
Reads as: Potential due to a dipole at large distance is one over four pi epsilon naught times p cos theta over r squared.
Variables: p: dipole moment (C·m); θ: angle between p and position vector; r: distance from dipole centre (m); r̂: unit vector along r.
Use when: Calculating potential due to an electric dipole for r >> a.
⚠ Students use this formula for points close to the dipole where the approximation fails.
V = (1/4πε₀) Σ (q_i / r_iP)
Reads as: Potential due to a system of point charges is one over four pi epsilon naught times sum of q_i over distance from each charge to point P.
Variables: q_i: i-th charge (C); r_iP: distance from q_i to point P (m).
Use when: Finding net potential from multiple discrete charges.
⚠ Students forget that potential is a scalar sum, not vector sum.
U = (1/4πε₀) (q₁q₂ / r₁₂)
Reads as: Potential energy of two point charges is one over four pi epsilon naught times product of charges over distance between them.
Variables: U: potential energy (J); q₁, q₂: charges (C); r₁₂: distance between charges (m).
Use when: Calculating electrostatic potential energy of a two-charge system.
⚠ Students forget that this is the energy of the system, not of a single charge.
U = (1/4πε₀) [ (q₁q₂/r₁₂) + (q₁q₃/r₁₃) + (q₂q₃/r₂₃) ]
Reads as: Potential energy of three charges is one over four pi epsilon naught times sum of pairwise products over distances.
Variables: Same as above for each pair.
Use when: Calculating potential energy for three point charges.
⚠ Students double-count pairs or miss a pair.
U = q V(r) (for a single charge in external field)
Reads as: Potential energy of charge q in external field is q times potential at that point.
Variables: U: potential energy (J); q: charge (C); V(r): external potential at position r (V).
Use when: Finding potential energy of a charge in an external field.
⚠ Students think this includes interaction with other charges in the system.
U = q₁ V(r₁) + q₂ V(r₂) + (1/4πε₀) (q₁q₂ / r₁₂)
Reads as: Potential energy of two charges in external field is sum of individual energies plus mutual interaction energy.
Variables: V(r₁), V(r₂): external potentials at positions of q₁ and q₂ (V); r₁₂: distance between charges (m).
Use when: Calculating total electrostatic energy of two charges in an external field.
⚠ Students forget to add the mutual interaction term.
U(θ) = –p·E = –pE cosθ
Reads as: Potential energy of a dipole in uniform field is negative of p dot E.
Variables: p: dipole moment (C·m); E: electric field (N/C); θ: angle between p and E.
Use when: Finding potential energy of a dipole in a uniform external electric field.
⚠ Students forget the negative sign and write U = pE cosθ.
τ = p × E
Reads as: Torque on a dipole is p cross E.
Variables: τ: torque (N·m); p: dipole moment (C·m); E: electric field (N/C).
Use when: Calculating torque experienced by a dipole in a uniform electric field.
⚠ Students confuse torque with potential energy.
|E| = – dV/dl
Reads as: Magnitude of electric field equals negative rate of change of potential with distance.
Variables: E: electric field (V/m); V: potential (V); l: distance along direction of field (m).
Use when: Relating electric field to potential gradient.
⚠ Students forget the negative sign and the direction: field points from high to low potential.
V = (1/4πε₀) (q/R) (for a spherical shell, r ≥ R)
Reads as: Potential outside a uniformly charged spherical shell is same as if all charge is at centre.
Variables: q: total charge on shell (C); R: radius of shell (m); r: distance from centre (m).
Use when: Finding potential outside or on the surface of a spherical shell.
⚠ Students use this formula for points inside the shell where potential is constant and equal to surface value.
How It Works
Work Done in Moving a Charge in an Electrostatic Field
  1. Identify the electrostatic force F_e on the test charge q due to the source charge configuration.
  2. Apply an external force F_ext exactly equal and opposite to F_e (F_ext = –F_e) to move the charge without acceleration.
  3. The work done by the external force W_RP = ∫_R^P F_ext · dl is stored as potential energy difference.
  4. The work done is independent of the path taken, depending only on initial and final positions.
💡 The external force must exactly cancel the electric force; otherwise, the charge accelerates and work done is not fully stored as potential energy.
Derivation of Potential Due to a Point Charge
  1. Consider a point charge Q at the origin. To find potential at distance r, bring a unit positive test charge from infinity to that point.
  2. Choose a radial path from infinity to P. At an intermediate point at distance r', the force on the test charge is (1/4πε₀)(Q/r'²) along r̂.
  3. Work done against this force over a small displacement dr' is dW = – (1/4πε₀)(Q/r'²) dr' (negative because dr' is negative when moving inward).
  4. Integrate from r' = ∞ to r' = r: W = ∫_∞^r – (1/4πε₀)(Q/r'²) dr' = (1/4πε₀)(Q/r).
  5. By definition, this work equals the potential V at P.
💡 The negative sign in the integrand is crucial; students often miss it and get the sign wrong.
Derivation of Potential Due to an Electric Dipole
  1. Place a dipole with charges +q and –q separated by 2a, centred at origin. Consider a point P at distance r from centre, making angle θ with dipole axis.
  2. Write distances r₁ (to +q) and r₂ (to –q) using cosine law: r₁² = r² + a² – 2ar cosθ, r₂² = r² + a² + 2ar cosθ.
  3. For r >> a, use binomial approximation: 1/r₁ ≈ (1/r)[1 + (a/r) cosθ] and 1/r₂ ≈ (1/r)[1 – (a/r) cosθ].
  4. Potential V = (1/4πε₀)[q/r₁ – q/r₂] ≈ (1/4πε₀)[ (2qa cosθ)/r² ] = (1/4πε₀)(p cosθ / r²).
💡 The approximation r >> a is essential; the formula is not valid for points close to the dipole.
Assembling a System of Charges (Potential Energy Calculation)
  1. Bring the first charge q₁ from infinity to its position. No work is done as there is no field initially.
  2. Bring the second charge q₂ from infinity to its position. Work done = q₂ × (potential due to q₁ at that point) = (1/4πε₀)(q₁q₂/r₁₂).
  3. For the third charge q₃, work done = q₃ × (potential due to q₁ and q₂ at its position) = (1/4πε₀)[q₁q₃/r₁₃ + q₂q₃/r₂₃].
  4. Total potential energy is the sum of work done in all steps: U = (1/4πε₀)[q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃].
💡 The order of bringing charges does not affect the final potential energy; it is path-independent.
Electrostatic Shielding and Properties of Conductors
  1. In a conductor, free electrons move until the net electric field inside becomes zero.
  2. By Gauss's law, if E = 0 inside, the net charge inside any closed surface is zero, so excess charge resides only on the surface.
  3. At the surface, the electric field must be perpendicular; otherwise, tangential component would cause current.
  4. Since E = 0 inside, potential is constant throughout the conductor and equal to its surface value.
💡 The field just outside a charged conductor is perpendicular to the surface and its magnitude is σ/ε₀, but this is not derived in this chapter.
Exam Traps — Don't Make These Mistakes
⚠ Sign of Work Done by External Force vs Electric Field
✗ Work done by external force equals work done by electric field.
→ The external force is opposite to the electric force (F_ext = –F_e), so work done by external force is negative of work done by electric field.
✓ Work done by external force = – (work done by electric field). The work done by external force is stored as potential energy.
⚠ Potential Energy of Two Charges: Sign Convention
✗ Potential energy of two opposite charges is positive because they attract.
→ For opposite charges, work must be done to separate them, so potential energy is negative (work done by external force to bring them together is negative).
✓ U = (1/4πε₀)(q₁q₂/r). For unlike charges (q₁q₂ < 0), U is negative. For like charges (q₁q₂ > 0), U is positive.
⚠ Potential Due to a Dipole in the Equatorial Plane
✗ Potential in the equatorial plane is maximum or minimum.
→ In the equatorial plane (θ = 90°), cosθ = 0, so V = 0. The potential is zero, not maximum or minimum.
✓ The potential due to a dipole in its equatorial plane is zero at all points (for r >> a).
⚠ Electric Field Inside a Conductor
✗ Electric field inside a charged conductor is non-zero because it has charge.
→ In electrostatic equilibrium, free charges redistribute to cancel any internal field. The field inside a conductor is always zero, regardless of whether it is charged or not.
✓ Inside a conductor in electrostatic equilibrium, the electric field is zero everywhere.
⚠ Potential Energy of a Dipole: Zero Reference
✗ Potential energy of a dipole is zero when it is aligned with the field (θ = 0°).
→ The natural choice for zero potential energy is when the dipole is perpendicular to the field (θ = 90°). At θ = 0°, U = –pE, which is minimum, not zero.
✓ U(θ) = –pE cosθ. Taking U = 0 at θ = 90°, then U(0°) = –pE and U(180°) = +pE.
⚠ Path Independence of Potential Difference
✗ Work done depends on the path taken because the force varies with distance.
→ Electrostatic force is conservative; work done depends only on initial and final positions, not on the path.
✓ Work done by electrostatic field in moving a charge between two points is independent of the path and depends only on the initial and final positions.
Compare & Contrast
Definition and dependence on charge
Electrostatic Potential (V)
Electrostatic Potential Energy (U)
Potential is work done per unit positive test charge (V = W/q), independent of the test charge. Potential energy is the work done on a specific charge q (U = qV), and depends on the magnitude of q.
Nature and distance dependence
Electric Field Due to a Point Charge
Electric Potential Due to a Point Charge
Electric field is a vector quantity and varies as 1/r². Electric potential is a scalar quantity and varies as 1/r.
Distance dependence and angular dependence
Potential Due to a Point Charge
Potential Due to a Dipole
Point charge potential falls off as 1/r and is spherically symmetric. Dipole potential falls off as 1/r² and depends on angle θ (V ∝ cosθ/r²).
Behaviour of charges and internal field
Conductor in Electrostatic Equilibrium
Insulator in Electrostatic Equilibrium
In a conductor, free charges move to make internal E = 0 and excess charge resides on the surface. In an insulator, charges are bound and cannot move; internal field may exist and charge remains where placed.
Exam Focus
Marks pattern: Typically 2 MCQs (1 mark each), 1 very short answer (1 mark), 1 short answer (2-3 marks), and 1 long answer (5 marks) — total around 10-12 marks.
Most asked: Potential due to a point charge and system of charges · Potential due to an electric dipole · Potential energy of a system of charges · Potential energy of a dipole in an external field · Properties of conductors (E = 0 inside, charge on surface, constant potential) · Relation between electric field and potential (E = –dV/dl) · Equipotential surfaces and their properties
Don't skip: Derivation of potential due to a dipole and comparison with point charge; Numerical on potential energy of a system of charges (e.g., four charges at corners of a square); Numerical on work done in moving a charge in an external field; Conceptual questions on properties of conductors and equipotential surfaces
One-Line Revision (read aloud before the exam)
  1. Electrostatic potential at a point is the work done per unit positive charge in bringing it from infinity to that point.
  2. Potential difference between two points is independent of the path taken.
  3. For a point charge Q, potential V = (1/4πε₀)(Q/r).
  4. For an electric dipole, potential V = (1/4πε₀)(p cosθ / r²) for r >> a.
  5. Potential on the dipole axis (θ = 0) is (1/4πε₀)(p/r²); in the equatorial plane (θ = 90°), it is zero.
  6. Equipotential surfaces are always perpendicular to electric field lines.
  7. No work is done in moving a charge on an equipotential surface.
  8. Inside a conductor in electrostatic equilibrium, the electric field is zero.
  9. Excess charge on a conductor resides only on its surface.
  10. Potential is constant throughout the volume of a conductor.
  11. Potential energy of two point charges U = (1/4πε₀)(q₁q₂/r₁₂).
  12. For like charges, U is positive; for unlike charges, U is negative.
  13. Potential energy of a dipole in a uniform field U = –p·E = –pE cosθ.
  14. Torque on a dipole in a uniform field τ = p × E.
  15. Electric field is the negative gradient of potential: E = –dV/dl.
  16. 1 eV = 1.6 × 10⁻¹⁹ J.
  17. The work done by an external force in assembling a charge configuration is stored as electrostatic potential energy.

MCQs — 10 Questions with Answers

Q1. The electrostatic potential at a distance r from a point charge Q is given by V = kQ/r. If Q is doubled and r is halved, the potential becomes:

  • A. 4 times the original ✓
  • B. 2 times the original
  • C. 8 times the original
  • D. remains the same

Answer: A — New potential V' = k(2Q)/(r/2) = 4kQ/r = 4V, so potential becomes 4 times the original.

Q2. A charge q is moved from point R to point P against an electric repulsive force. Which statement is correct?

  • A. Work done by external force is negative and stored as kinetic energy
  • B. Work done by external force is positive and stored as potential energy ✓
  • C. Work done by electric force is positive and equals potential energy gained
  • D. Work done by external force equals work done by electric force

Answer: B — Moving against repulsive force requires positive work by external force, which is stored as potential energy; work by electric force is negative.

Q3. Why is the test charge used to measure electric potential assumed to be infinitesimally small?

  • A. To reduce experimental error in measurement
  • B. To ensure it does not disturb the original charge configuration creating the field ✓
  • C. To make calculations simpler using small angle approximation
  • D. Because only small charges experience electric forces

Answer: B — An infinitesimal test charge ensures the measured potential is characteristic of the original field only, not altered by the test charge itself.

Q4. The work done in moving a +2 C charge from potential 5 V to potential 8 V is:

  • A. 6 J
  • B. +6 J ✓
  • C. −6 J
  • D. 10 J

Answer: B — Work done by external force = q(VP − VR) = 2 × (8 − 5) = 2 × 3 = +6 J (positive work required to move positive charge to higher potential).

Q5. Which of the following is NOT correct about electrostatic potential in a conservative field?

  • A. Potential difference is path-independent
  • B. Potential energy can be added an arbitrary constant without changing physics
  • C. Potential is a vector quantity like electric field ✓
  • D. Potential at infinity is conventionally taken as zero

Answer: C — Potential V is a scalar quantity (work per unit charge), not a vector; electric field E is the vector quantity in electrostatics.

Q6. A charge +q is brought slowly from infinity to a point where potential is V. The external force applied is:

  • A. Always equal to electric force in magnitude and direction
  • B. Equal in magnitude but opposite in direction to electric force at each point ✓
  • C. Zero, because we consider infinitesimal test charge
  • D. Greater than electric force to accelerate the charge

Answer: B — To move the charge slowly (without acceleration), external force must exactly balance electric force at each point, equal in magnitude but opposite in direction.

Q7. Two identical charges are placed 1 m apart. The potential at the midpoint due to both charges is 200 V. If the charges are brought to 0.5 m apart, the potential at the midpoint becomes:

  • A. 200 V
  • B. 400 V
  • C. 800 V ✓
  • D. 100 V

Answer: C — At 1 m apart, each charge is 0.5 m from midpoint: V_total = 2(kq/0.5) = 200 V. At 0.5 m apart, each charge is 0.25 m from midpoint: V_total = 2(kq/0.25) = 8(kq/0.5) = 800 V.

Q8. Consider two statements: (A) Potential energy of a system has an arbitrary additive constant. (B) Only potential energy difference is physically meaningful. Choose correct option:

  • A. Both A and B are correct ✓
  • B. A is correct, B is incorrect
  • C. A is incorrect, B is correct
  • D. Both A and B are incorrect

Answer: A — Statement A is true because adding constant α to all points gives same PE difference; Statement B is true because physics depends only on ΔU, not absolute U values.

Q9. The electric potential at a distance r from a uniformly charged sphere of charge Q is V = kQ/r for r ≥ R. Work done in bringing a charge −2q from infinity to distance 2R from centre is:

  • A. −2kqQ/R
  • B. −kqQ/R ✓
  • C. −kqQ/(2R)
  • D. kqQ/R

Answer: B — W = q(V_final − V_initial) = (−2q)[kQ/(2R) − 0] = −kqQ/R; potential at 2R is kQ/(2R), and work is negative because negative charge moves against field.

Q10. A charge is moved along a closed path in an electric field and returns to its starting point. The work done by the electric force is:

  • A. Positive and maximum
  • B. Negative and minimum
  • C. Zero in a conservative field ✓
  • D. Depends on the shape of the path

Answer: C — In a conservative field, work done over a closed path is zero because initial and final positions are identical, so ΔU = 0, hence W_electric = 0.

Flashcards

What is electrostatic potential energy of charge q at a point?

The work done by an external force in bringing charge q from infinity to that point without acceleration.

Define electrostatic potential V at a point in an electric field.

The work done per unit positive charge in bringing it from infinity to that point without acceleration.

Why is Coulomb force called a conservative force?

Because the work done by Coulomb force between two charges depends only on initial and final positions, not on the path taken.

What is the relationship between potential energy U, charge q, and potential V?

U = qV, where U is potential energy at a point with potential V for a charge q.

Why do we choose potential energy to be zero at infinity?

This is a convenient reference point that removes the arbitrary additive constant and makes potential energy unique at every point.

What is the SI unit of electrostatic potential?

Volt (V), defined as joule per coulomb (J/C).

Explain why the test charge used to measure potential must be infinitesimally small.

An infinitesimal test charge does not disturb the original charge configuration and ensures the measured potential is characteristic of the field only.

What does path-independence of work in electrostatic field imply?

The work done depends only on the initial and final points, making the concept of potential energy meaningful and allowing us to define potential as a scalar field.

How is potential energy related to kinetic energy in a conservative field?

The total mechanical energy (KE + PE) is conserved; when external force is removed, PE converts to KE as the charge moves.

Why is external force applied 'equal and opposite' to electric force when moving a charge slowly?

This ensures zero net force and zero acceleration, allowing work done by external force to be stored entirely as potential energy without giving kinetic energy.

Important Board Questions

Define electrostatic potential at a point. State its SI unit. How is it different from electric field? [2 marks]

Define V as work per unit positive charge from ∞; give SI unit (Volt); state V is scalar while E is vector.

Explain with full working why electrostatic potential energy difference between two points is path-independent. Show using definition that ΔU depends only on initial and final positions, not path taken. Why is this property essential for defining potential energy? [5 marks]

Use definition ΔU = W_external from R to P; argue that in Coulomb force (inverse-square), circulation integral around closed loop = 0, proving path-independence; state this is characteristic of conservative forces, allowing scalar potential concept.

Derive the relationship U = qV between potential energy, charge, and potential. Starting from the definition of work done in moving charge q from infinity to point P, show step-by-step how potential is defined per unit charge. Then apply this to explain why a positive charge spontaneously moves from higher to lower potential region. What role does the arbitrary zero reference play? [6 marks]

Start with W_∞P = U_P for charge q; divide by q to get V_P = U_P/q; show positive charge moves to lower PE (lower V) because electric force does positive work; discuss why setting V_∞ = 0 removes arbitrary constant and ensures unique potential everywhere.

Assertion-Reason Questions for Electrostatic Potential and Capacitance

Q1. Assertion (A):

The electrostatic potential due to a point charge Q at distance r is always positive when Q > 0.

Reason (R):

Potential is defined as work done per unit positive charge, and bringing a positive test charge toward a positive source charge requires external work against repulsion.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both statements are true. When Q > 0, the potential V = kQ/r is positive for all finite r > 0. The reason correctly explains why: external work must be done to bring a positive test charge from infinity against the repulsive force, and this work stored as potential is positive. The assertion, reason, and their relationship are all correct.

Q2. Assertion (A):

In a uniform electric field, the equipotential surfaces are planes perpendicular to the electric field direction.

Reason (R):

Equipotential surfaces are defined as surfaces where the potential difference between any two points is zero.

Answer: C — A is true but R is false — The assertion is true—in uniform fields, equipotentials are indeed perpendicular planes to E. However, the reason is false. Equipotential surfaces are surfaces where potential is constant everywhere, not where potential difference is zero (that would be the same point). Both assertion and reason are true, but the reason does not explain why the surfaces are perpendicular; it only defines what equipotentials are.

Q3. Assertion (A):

The potential at the surface of an isolated conducting sphere of radius R carrying charge Q is V = kQ/R.

Reason (R):

Because the electric field inside the conductor is zero, all the charge resides on the surface and creates the same potential as a point charge at the center.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are correct, and the reason properly explains the assertion. The conductor reaches electrostatic equilibrium with E = 0 inside, all charge on surface, and potential determined by the total charge and radius as if it were concentrated at the center. This is a direct consequence of Gauss's law applied to the conductor.

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Case-Based Questions (CBQ) for Electrostatic Potential and Capacitance

Case Study:

A Van de Graaff generator in a physics laboratory accumulates charge on a conducting sphere of radius 0.5 m. When fully charged, the potential of the sphere is measured to be 150,000 V relative to the ground (taken as zero potential at infinity). The operator wants to calculate the total charge on the sphere and the electric potential at a distance of 2 m from the center of the sphere. Assume the sphere is isolated and the air around it has permittivity ε₀.

Q1. The total charge Q on the sphere is closest to (use k = 9 × 10⁹ N⋅m²/C²):

  • A. 8.3 × 10⁻⁶ C ✓
  • B. 8.3 × 10⁻³ C
  • C. 8.3 C
  • D. 83 C

Answer: A

Q2. The electric potential at 2 m from the center is:

  • A. 150,000 V
  • B. 75,000 V
  • C. 37,500 V ✓
  • D. 6,250 V

Answer: C

Q3. If a small conducting sphere (charge 10 μC) is placed near the Van de Graaff sphere at 2 m distance, the potential energy of this small sphere in the field is approximately:

  • A. 0.375 J ✓
  • B. 3.75 J
  • C. 37.5 J
  • D. 0.0375 J

Answer: A

Q4. As the operator charges the Van de Graaff sphere further, which statement is true?

  • A. Both the charge and potential increase, but potential increases faster (non-linearly)
  • B. The potential increases linearly with charge at fixed radius ✓
  • C. The charge increases while potential remains constant due to grounding
  • D. The ratio of potential to charge depends on the distance from the sphere

Answer: B

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Previous Year Questions (PYQ) for Electrostatic Potential and Capacitance

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 55/1/1 · 1 mark

In a region, the electric potential varies as V = 10 − 50x, where V is in volts and x in meters. The electric field in the region is

  • 10 N/C along +x
  • 10 N/C along −x
  • 50 N/C along +x
  • 50 N/C along −x

CBSE 2026 · Set 55/1/1 · 3 marks

A parallel plate capacitor of capacitance C has a dielectric slab between its plates. It is charged to a potential difference V by connecting it across a battery. The battery is then disconnected. If the dielectric slab is now withdrawn from the capacitor, how will the following be affected? (a) Capacitance of the capacitor, (b) Energy stored in the capacitor, and (c) The potential difference between the plates of the capacitor. Justify your answer in each case.

CBSE 2026 · Set 55/1/2 · 1 mark

In a region electric field is given by E⃗ = 4 xî N/C. The potential difference between points A (x = 1 m) and B (x = 3 m), (V_A − V_B) is

  • −16 V
  • 16 V
  • − 8 V
  • 8 V

CBSE 2026 · Set 55/1/2 · 3 marks

A 12.0 μF capacitor is charged to a potential difference of 150 V. The terminals of the charged capacitor are then connected to those of an uncharged 6.0 μF capacitor. Calculate final potential difference across and charge on, each capacitor.

CBSE 2026 · Set 55/1/3 · 1 mark

The electric potential for various points in x-y plane is given by V = 1.0 x² − 2.0 y², where V is in volts and x, y are in metres. The angle that the electric field at point (2.0 m, 1.0 m) makes with the positive x-axis is

  • 45°
  • 90°
  • 135°
  • 315°

CBSE 2026 · Set 55/1/3 · 3 marks

A parallel plate capacitor of capacitance C is charged to V volt by a battery. After sometime the battery is disconnected and the distance between the plates is doubled. A slab of dielectric constant k = 1.8 is then introduced to completely fill the space between the plates. How will the following be affected? (a) The capacitance of the capacitor. (b) The electric field between the plates of the capacitor. (c) The energy stored in the capacitor. Justify your answer in each case.

CBSE 2026 · Set 55/2/1 · 1 mark

Two metal spheres of radii r₁ and r₂ (> r₁) having charges q₁ and q₂ respectively kept in air, are brought in contact. Which of the following statements is not correct?

  • The total charge of the two spheres is conserved.
  • Both spheres attain the same potential.
  • The final potential of the system equals (1/(4πε₀)) (q₁ + q₂)/(r₁ + r₂)
  • The final potential of the system equals (1/(4πε₀)) (q₁ + q₂)(r₁ + r₂)/(r₁r₂)

CBSE 2026 · Set 55/2/1 · 5 marks

(a) Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates. (b) Two air-filled capacitors of capacitances C₁ and C₂ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.

CBSE 2026 · Set 55/2/3 · 5 marks

(a) Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

CBSE 2026 · Set 55(B)_Physics (Theory) · 1 mark

When an electric dipole of charges q and dipole moment p is placed in a uniform electric field E making an angle θ with the electric field E, then two charges experience equal and opposite forces +qE and –qE. These equal and opposite forces cancel each other. As the two forces are not collinear so a torque acts on the dipole. Since the net force acting on the dipole is zero, no work is done in moving the electric dipole in a uniform electric field, but some work will be done in rotating the dipole against the torque acting on it.

An electric dipole having dipole moment of 4 × 10⁻⁹ Cm is placed in a uniform electric field of 5 × 10⁴ N/C. If the angle between the dipole moment and the electric field is 30° then the value of potential energy of the dipole is –

  • 3.23 × 10⁻⁴ J
  • 1.0 × 10⁻⁵ J
  • 1.73 × 10⁻⁴ J
  • 5 × 10⁴ J

CBSE 2026 · Set 55/3/1 · 1 mark

Capacitors are manufactured with certain standard capacitances and working voltages. However, these standard values may not be the ones that are actually needed in a particular application. Two or more capacitors can be grouped in series or in parallel to achieve desired capacitance and voltage. When connected in series, the total capacitance decreases while the voltage rating increases, whereas in parallel connections, the total capacitance increases and maintains the same voltage rating. A capacitor stores energy in the electric field between its plates and stored energy is proportional to the square of the voltage and capacitance U = (1/2)CV2, where symbols have their usual meanings.

Two capacitors, one of 3 μF and the other of 6 μF, are connected in series in the circuit. The total capacitance of the circuit is:

  • 6 μF
  • 3 μF
  • 9 μF
  • 2 μF

CBSE 2026 · Set 55/3/1 · 5 marks

(a) (ii) Two point charges − 2 μC and 5 μC are placed at (− 30 cm, 0) and (30 cm, 0) respectively in an external electric field E⃗ = (A/x²) î, where A = 9 × 10⁵ Nm² C⁻¹. Find the electrostatic potential energy of this configuration.

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Class 12 Physics Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Physics Previous Year Questions, Chapter-wise

All 962 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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Class 12 Physics Derivations

All 68 derivations, chapter-wise, step by step with diagrams and marked with the CBSE board years (2022–2026) each was asked in. Free PDF download.

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