Class 12 · Physics · Part 1

Chapter 2: Electrostatic Potential and Capacitance

Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.

2.1Electrostatic potentialWork done on a charge, potential difference and the link with the field

Potential energy and potential

The electrostatic force is conservative: the work done in moving a charge between two points does not depend on the path taken.
The potential difference between two points is the work done per unit positive charge in moving it from one point to the other without acceleration: V_P − V_R = W_RP / q.
The potential at a point is the work done in bringing a unit positive charge from infinity to that point. (The potential at infinity is taken as zero.)
Unit: volt (V), where 1 V = 1 J C⁻¹. Potential is a scalar.
The work done by an external agent in moving a charge q from A to B is W = q (V_B − V_A).

Potential and the field

The electric field points in the direction in which the potential falls most steeply. In a uniform field E = −ΔV / Δl, so E can be measured in V m⁻¹ (same as N C⁻¹).
The work done by the field in moving a charge q from a point at V₁ to a point at V₂ is q (V₁ − V₂).
A positive charge, left free, moves from higher to lower potential; a negative charge moves from lower to higher potential.
A useful unit of energy: 1 electron volt (eV) is the energy gained by an electron moving through 1 V: 1 eV = 1.6 × 10⁻¹⁹ J.

2.2Potential due to a point chargeV = kq/r and potential of a system of charges

Point charge

At a distance r from a point charge q, V = k q / r = q / (4π ε₀ r).
The potential is positive near a positive charge and negative near a negative charge. It falls as 1/r, slower than the field, which falls as 1/r².
Points at the same distance from a point charge are at the same potential.

System of charges

Potential is a scalar, so the potential at a point due to several charges is the algebraic sum: V = k (q₁/r₁ + q₂/r₂ + q₃/r₃ + …).
Include the sign of each charge. Unlike the field, no resolving into components is needed.

2.3Potential due to an electric dipoleAxial and equatorial points

Dipole potential

At a point at distance r from the centre of a dipole, making angle θ with the dipole moment, for r ≫ a: V = k p cos θ / r².
On the axis (θ = 0°): V = k p / r², positive on the side of the positive charge and negative on the other side.
On the equatorial line (θ = 90°): V = 0.
The dipole potential falls as 1/r² (a point charge: 1/r).

2.4Equipotential surfacesSurfaces of constant potential

Properties

An equipotential surface is a surface on which the potential has the same value at every point.
No work is done in moving a charge along an equipotential surface.
The electric field is always perpendicular to the equipotential surface at every point, and points from higher to lower potential.
For a point charge the surfaces are concentric spheres; in a uniform field they are parallel planes perpendicular to the field.
Two equipotential surfaces never intersect, because a point cannot have two potentials.
The surfaces are closer together where the field is stronger.

Equipotential surfaces (dashed) are always perpendicular to the field lines. Moving along one needs no work.

2.5Potential energy of chargesEnergy of a system and of a charge in an external field

Potential energy of a system of charges

The potential energy of a system is the work done in assembling the charges, bringing each one in from infinity.
For two charges: U = k q₁ q₂ / r₁₂. The energy is positive for like charges and negative for unlike charges.
For three charges: U = k (q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃). Each pair is counted once.

In an external field

A charge q at a point where the external potential is V has potential energy U = q V.
For a dipole in a uniform field, U = −p E cos θ. It is lowest (stable) at θ = 0° and highest at θ = 180°.
The work done in turning a dipole from θ₁ to θ₂ is W = pE (cos θ₁ − cos θ₂).

2.6Electrostatics of conductorsFields and charges in and on conductors

Properties of a conductor in electrostatic equilibrium

The field inside a conductor is zero.
At the surface the field is perpendicular to the surface, with magnitude E = σ / ε₀.
Any excess charge resides on the surface; there is no net charge in the interior.
The potential is the same everywhere inside and on the surface (the whole conductor is equipotential).
The field is stronger where the surface is more sharply curved, where σ is larger.

Cavities and shielding

The field inside an empty cavity of a charged conductor is zero, whatever the charge outside.
This is electrostatic shielding: a metal box protects instruments from external electric fields. This is why it is safe in a car in a thunderstorm.

2.7Dielectrics and polarisationInsulators in an electric field

Polar and non-polar molecules

In a non-polar molecule (such as O₂, H₂) the centres of positive and negative charge coincide. An external field pulls them apart and creates an induced dipole moment.
In a polar molecule (such as H₂O, HCl) the centres do not coincide, so each molecule has a permanent dipole moment. Without a field they point randomly; a field partly aligns them.

Polarisation

In both cases the dielectric develops a net dipole moment per unit volume in the direction of the field. This is polarisation.
The slab acquires induced surface charge, negative on the face near the positive plate and positive on the face near the negative plate.
The induced charge produces a field opposite to the applied field, so the net field inside is reduced: E = E₀ / K, where K is the dielectric constant of the material.

Polarisation of a dielectric in a capacitor. The induced charges on the slab faces oppose the applied field, reducing it to E₀/K.

2.8Capacitors and capacitanceStoring charge

Capacitor

A capacitor is a system of two conductors, separated by an insulator, that stores charge and energy. The charges on the two conductors are +Q and −Q.
Capacitance: C = Q / V. The unit is the farad (F): 1 F = 1 C V⁻¹. Common units are μF = 10⁻⁶ F, nF = 10⁻⁹ F, pF = 10⁻¹² F.
Capacitance depends only on the geometry (size, shape, separation) and the medium between the plates, not on Q or V.

2.9The parallel plate capacitorC = ε₀A/d

Capacitance of a parallel plate capacitor

Two plates of area A separated by a small distance d carry charge +Q and −Q. The field between them is E = σ/ε₀ = Q / (ε₀ A).
The potential difference is V = E d = Q d / (ε₀ A), so C = ε₀ A / d.
Larger plate area or smaller separation gives larger capacitance.
The edge effects (fringing) are neglected when d is small compared with the size of the plates.

A parallel plate capacitor. The field is uniform between the plates (and bulges only at the edges), so E = σ/ε₀ = Q/(ε₀A).

2.10Effect of a dielectricCapacitance with a material between the plates

Dielectric filling the gap

With a dielectric of constant K filling the space between the plates, C = K ε₀ A / d = K C₀, where C₀ is the capacitance without it.
The dielectric constant K is always greater than 1 (for a vacuum K = 1).

Slab of thickness t

If a slab of dielectric constant K and thickness t (less than d) is inserted, C = ε₀ A / (d − t + t/K).
For a conducting slab the field inside is zero, equivalent to K → ∞: C = ε₀ A / (d − t).

Battery connected or not

Battery connected (V constant): the capacitance increases by K, so Q = CV increases by K, and the energy ½CV² increases by K.
Battery disconnected (Q constant): V falls to V₀/K, the field falls to E₀/K, and the energy Q²/2C falls to 1/K of its value.

2.11Combination of capacitorsSeries and parallel

Series

Capacitors joined end to end carry the same charge Q; the voltages add: V = V₁ + V₂ + V₃.
1/C = 1/C₁ + 1/C₂ + 1/C₃. The combination is smaller than the smallest capacitor.
For two capacitors, C = C₁C₂ / (C₁ + C₂). For n equal capacitors C, the total is C/n.
The largest voltage falls across the smallest capacitor.

Parallel

Capacitors joined between the same two points have the same voltage V; the charges add: Q = Q₁ + Q₂ + Q₃.
C = C₁ + C₂ + C₃. The combination is larger than any one of them.

Combinations of capacitors. In series the charge is the same; in parallel the voltage is the same.

2.12Energy stored in a capacitorWhere the energy is

Energy

The work done in charging a capacitor is stored as electrostatic energy: U = ½ Q V = ½ C V² = Q² / 2C.
Energy density (energy per unit volume of the field): u = ½ ε₀ E². The energy is stored in the field between the plates.
When a charged capacitor is connected to an uncharged one, charge is shared and some energy is lost (as heat and radiation) although the charge is conserved.
Capacitors are used in the flash of a camera, in smoothing power supplies and in tuning circuits.

Quick referenceFormulas at a glanceRevise before the exam

Key results

Potential: V = kq/r; system: V = k Σ qᵢ/rᵢ
Potential energy of two charges: U = k q₁q₂ / r
Dipole: V = kp cos θ / r²; U = −pE cos θ
Capacitance: C = Q/V; plates: C = ε₀A/d; with dielectric: C = Kε₀A/d
Series: 1/C = Σ 1/Cᵢ; parallel: C = Σ Cᵢ
Energy: U = ½CV² = Q²/2C = ½QV; density ½ε₀E²
Conductor surface: E = σ/ε₀, inside E = 0

Common mistakes

Adding potentials as vectors, or forgetting the signs of the charges.
Using 1/C = C₁ + C₂ for the parallel combination (or the reverse).
Forgetting whether the battery is connected (V fixed) or removed (Q fixed) when a dielectric is inserted.
Writing the capacitance in pF but substituting it as F in the energy formula.

Try each problem first, then open the solution to check your working.

2.1PotentialPotential, potential difference and work · 8 problems

Q1Potential of a point charge

Find the potential at a distance of 20 cm from a point charge of +4 μC.

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Given: q = 4 × 10⁻⁶ C; r = 0.2 m

Use: V = k q / r

V = (9 × 10⁹ × 4 × 10⁻⁶) / 0.2 = 3.6 × 10⁴ / 0.2 = 1.8 × 10⁵ V

Answer: 1.8 × 10⁵ V

Q2Potential of two charges

Charges +3 μC and −5 μC are at distances 10 cm and 20 cm from a point P. Find the potential at P.

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Given: q₁ = +3 μC at 0.1 m; q₂ = −5 μC at 0.2 m

Use: Add the potentials with their signs: V = k (q₁/r₁ + q₂/r₂)

V = 9 × 10⁹ × (3 × 10⁻⁶ / 0.1 − 5 × 10⁻⁶ / 0.2)

= 9 × 10⁹ × (3 × 10⁻⁵ − 2.5 × 10⁻⁵) = 9 × 10⁹ × 5 × 10⁻⁶ = 4.5 × 10⁴ V

Answer: 4.5 × 10⁴ V

Q3Potential at the centre of a square

Four charges +2 μC, −2 μC, +2 μC and −2 μC are placed at the corners of a square of side 20 cm in that order. Find the potential and the field at the centre.

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Given: q = ±2 μC; each corner is 0.1√2 m = 0.141 m from the centre

Use: Potential is a scalar: add the values

The two positive charges are on one diagonal and the two negative charges on the other, so every charge is at the same distance from the centre.

Potential (scalar): V = k(+2 + 2 − 2 − 2) × 10⁻⁶ / 0.141 = 0.

Field (vector): the two + charges are at opposite ends of a diagonal, so their fields at the centre are equal and opposite and cancel. The same happens for the two − charges.

So the net field at the centre is also zero.

Answer: V = 0 and E = 0 at the centre

Q4Work done in moving a charge

Find the work done in moving a charge of +2 μC from a point at 100 V to a point at 400 V.

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Given: q = 2 × 10⁻⁶ C; V_A = 100 V; V_B = 400 V

Use: W = q (V_B − V_A)

W = 2 × 10⁻⁶ × (400 − 100) = 6 × 10⁻⁴ J

Answer: 6 × 10⁻⁴ J (done by the external agent)

Q5Electron through a potential difference

An electron is accelerated from rest through a potential difference of 100 V. Find its kinetic energy in eV and its speed. (mₑ = 9.11 × 10⁻³¹ kg)

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Given: V = 100 V; e = 1.6 × 10⁻¹⁹ C

Use: K = eV and ½mv² = eV

K = 100 eV = 100 × 1.6 × 10⁻¹⁹ = 1.6 × 10⁻¹⁷ J

v = √(2K/m) = √(2 × 1.6 × 10⁻¹⁷ / 9.11 × 10⁻³¹) = 5.93 × 10⁶ m/s

Answer: 100 eV; v ≈ 5.9 × 10⁶ m/s

Q6Field from potential

The potential difference between two parallel plates 2 cm apart is 500 V. Find the field between them, assuming it is uniform.

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Given: ΔV = 500 V; d = 0.02 m

Use: E = ΔV / d

E = 500 / 0.02 = 2.5 × 10⁴ V/m

Answer: 2.5 × 10⁴ V/m (N/C), directed from the higher to the lower potential

Q7Potential of a charged sphere

A metal sphere of radius 10 cm carries a charge of 5 μC. Find the potential (a) at its surface, (b) at its centre and (c) at 30 cm from the centre.

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Given: q = 5 × 10⁻⁶ C; R = 0.1 m

Use: Outside and on the surface the sphere acts like a point charge at its centre; inside, the potential is constant

(a) V = kq/R = 9 × 10⁹ × 5 × 10⁻⁶ / 0.1 = 4.5 × 10⁵ V

(b) Inside a conductor the potential equals that at the surface: 4.5 × 10⁵ V

Answer: (a) 4.5 × 10⁵ V; (b) 4.5 × 10⁵ V; (c) 1.5 × 10⁵ V

Q8Zero potential on a line

Charges +4 μC and −1 μC are 30 cm apart. Find the points on the line joining them where the potential is zero.

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Given: q₁ = 4 μC, q₂ = −1 μC, d = 0.3 m

Use: V = 0 where q₁/r₁ + q₂/r₂ = 0

Between them: 4/x = 1/(0.3 − x), so 4(0.3 − x) = x, 1.2 = 5x, x = 0.24 m from the 4 μC charge.

Outside, on the side of the smaller charge: 4/x = 1/(x − 0.3), so 4x − 1.2 = x, x = 0.4 m from the 4 μC charge.

Answer: 0.24 m and 0.4 m from the 4 μC charge

2.2Potential energySystems of charges and dipoles · 5 problems

Q9Energy of two charges

Two point charges +3 μC and +2 μC are 10 cm apart. Find the potential energy of the pair.

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Given: q₁ = 3 μC; q₂ = 2 μC; r = 0.1 m

Use: U = k q₁ q₂ / r

U = (9 × 10⁹ × 3 × 10⁻⁶ × 2 × 10⁻⁶) / 0.1 = 0.054 / 0.1 = 0.54 J

Answer: 0.54 J (positive, since the charges are alike)

Q10Three charges at the corners of a triangle

Three charges of +1 μC each are at the corners of an equilateral triangle of side 10 cm. Find the potential energy of the system.

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Given: q = 10⁻⁶ C; side = 0.1 m

Use: Sum over the three pairs: U = 3 × k q² / a

U = 3 × (9 × 10⁹ × 10⁻¹²) / 0.1 = 3 × 0.09 = 0.27 J

Answer: 0.27 J

Q11Potential energy of a dipole

A dipole of moment 2 × 10⁻⁸ C m is in a uniform field of 10⁴ N/C. Find its potential energy when p is (a) along E, (b) perpendicular to E and (c) opposite to E.

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Given: p = 2 × 10⁻⁸ C m; E = 10⁴ N/C

Use: U = −pE cos θ

pE = 2 × 10⁻⁸ × 10⁴ = 2 × 10⁻⁴ J

(a) θ = 0°: U = −2 × 10⁻⁴ J

(b) θ = 90°: U = 0

Answer: (a) −2 × 10⁻⁴ J; (b) 0; (c) +2 × 10⁻⁴ J

Q12Work in turning a dipole

How much work is needed to turn the dipole of the previous problem from θ = 0° to θ = 60°?

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Given: pE = 2 × 10⁻⁴ J; θ₁ = 0°; θ₂ = 60°

Use: W = pE (cos θ₁ − cos θ₂)

W = 2 × 10⁻⁴ × (1 − 0.5) = 1 × 10⁻⁴ J

Answer: 1 × 10⁻⁴ J

Q13Dipole potential on the axis

A short dipole of moment 4 × 10⁻⁹ C m is placed in air. Find the potential at a point on its axis 0.3 m from the centre, on the side of the positive charge, and at a point on the equatorial line.

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Given: p = 4 × 10⁻⁹ C m; r = 0.3 m

Use: V = k p cos θ / r²; on the equatorial line cos 90° = 0

On the axis (θ = 0°): V = (9 × 10⁹ × 4 × 10⁻⁹) / (0.3)² = 36 / 0.09 = 400 V

On the equatorial line: V = 0

Answer: 400 V on the axis (positive side); 0 on the equatorial line

2.3CapacitanceParallel plates and dielectrics · 8 problems

Q14Parallel plate capacitor

A parallel plate capacitor has plates of area 200 cm² separated by 1 mm in air. Find its capacitance. (ε₀ = 8.85 × 10⁻¹² F/m)

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Given: A = 200 cm² = 0.02 m²; d = 10⁻³ m

Use: C = ε₀ A / d

C = (8.85 × 10⁻¹² × 0.02) / 10⁻³ = 1.77 × 10⁻¹⁰ F

Answer: 1.77 × 10⁻¹⁰ F (177 pF)

Q15Charge and voltage

A 10 μF capacitor is charged to 50 V. Find the charge on it.

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Given: C = 10 × 10⁻⁶ F; V = 50 V

Use: Q = C V

Q = 10 × 10⁻⁶ × 50 = 5 × 10⁻⁴ C

Answer: 5 × 10⁻⁴ C (500 μC)

Q16Dielectric and spacing

The capacitance of an air capacitor is 5 pF. Find the new capacitance when the gap is halved and filled with a material of dielectric constant 4.

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Given: C₀ = 5 pF; d → d/2; K = 4

Use: C = K ε₀ A / d

Halving d doubles C, and the dielectric multiplies it by 4: C = 5 × 2 × 4 = 40 pF

Answer: 40 pF

Q17Partly filled capacitor

The plates of a capacitor have area 200 cm² and are 1 mm apart. A 0.5 mm thick slab of dielectric constant 5 is slipped in between them. Find the capacitance.

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Given: A = 0.02 m²; d = 1 mm; t = 0.5 mm; K = 5

Use: C = ε₀ A / (d − t + t/K)

d − t + t/K = 1 − 0.5 + 0.5/5 = 0.6 mm = 6 × 10⁻⁴ m

C = (8.85 × 10⁻¹² × 0.02) / (6 × 10⁻⁴) = 2.95 × 10⁻¹⁰ F

Answer: 2.95 × 10⁻¹⁰ F (295 pF)

Q18Conducting slab

Find the capacitance if the slab inserted in the previous problem is a 0.5 mm thick sheet of metal.

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Given: A = 0.02 m²; d = 1 mm; t = 0.5 mm

Use: C = ε₀ A / (d − t)

C = (8.85 × 10⁻¹² × 0.02) / (0.5 × 10⁻³) = 3.54 × 10⁻¹⁰ F

Answer: 3.54 × 10⁻¹⁰ F (354 pF)

Q19Plate separation

What separation is needed between two plates, each of area 100 cm², to make a 17.7 pF air capacitor?

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Given: A = 0.01 m²; C = 17.7 × 10⁻¹² F

Use: d = ε₀ A / C

d = (8.85 × 10⁻¹² × 0.01) / (17.7 × 10⁻¹²) = 5 × 10⁻³ m

Answer: 5 mm

Q20Field and charge on the plates

A parallel plate capacitor with plates 4 mm apart is connected to a 200 V battery. Find the electric field between the plates and the surface charge density on the plates.

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Given: V = 200 V; d = 4 × 10⁻³ m

Use: E = V/d; σ = ε₀ E

E = 200 / (4 × 10⁻³) = 5 × 10⁴ V/m

σ = 8.85 × 10⁻¹² × 5 × 10⁴ = 4.4 × 10⁻⁷ C/m²

Answer: E = 5 × 10⁴ V/m; σ = 4.4 × 10⁻⁷ C/m²

Q21Capacitance of an isolated sphere

Find the capacitance of an isolated conducting sphere of radius 10 cm. (C = 4πε₀R)

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Given: R = 0.1 m

Use: C = 4π ε₀ R

C = 4π × 8.85 × 10⁻¹² × 0.1 = 1.11 × 10⁻¹¹ F

Answer: 1.1 × 10⁻¹¹ F (11 pF)

2.4Combinations of capacitorsSeries, parallel and mixed · 6 problems

Q22Series combination

Capacitors of 2 μF, 3 μF and 6 μF are connected in series. Find the equivalent capacitance.

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Given: C₁ = 2, C₂ = 3, C₃ = 6 μF

Use: 1/C = 1/C₁ + 1/C₂ + 1/C₃

1/C = 1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = 1

C = 1 μF

Answer: 1 μF

Q23Parallel combination

Find the equivalent capacitance when the three capacitors of the previous problem are connected in parallel.

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Given: C₁ = 2, C₂ = 3, C₃ = 6 μF

Use: C = C₁ + C₂ + C₃

C = 2 + 3 + 6 = 11 μF

Answer: 11 μF

Q24Voltage across series capacitors

The 2 μF, 3 μF and 6 μF capacitors are connected in series to a 12 V battery. Find the charge on each and the voltage across each.

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Given: Ceq = 1 μF; V = 12 V

Use: Series: the same charge on each; V = Q/C

Q = Ceq V = 1 × 10⁻⁶ × 12 = 1.2 × 10⁻⁵ C = 12 μC on each

V₁ = 12 / 2 = 6 V; V₂ = 12 / 3 = 4 V; V₃ = 12 / 6 = 2 V

Check: 6 + 4 + 2 = 12 V.

Answer: 12 μC on each; 6 V, 4 V and 2 V

Q25Charge on parallel capacitors

The 2 μF, 3 μF and 6 μF capacitors are connected in parallel across a 12 V battery. Find the charge on each.

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Given: V = 12 V

Use: Parallel: same voltage; Q = CV

Q₁ = 2 × 12 = 24 μC; Q₂ = 3 × 12 = 36 μC; Q₃ = 6 × 12 = 72 μC

Total = 132 μC = 11 μF × 12 V ✓

Answer: 24 μC, 36 μC and 72 μC

Q26A mixed network

Two 6 μF capacitors in parallel are connected in series with a 3 μF capacitor. Find the total capacitance.

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Given: Parallel pair: 6 + 6 = 12 μF; then in series with 3 μF

Use: Combine step by step

Parallel pair: 6 + 6 = 12 μF

Series with 3 μF: C = (12 × 3) / (12 + 3) = 36 / 15 = 2.4 μF

Answer: 2.4 μF

Q27Capacitors in a ratio

A 4 μF and a 12 μF capacitor are connected in series across a 100 V supply. Find the voltage across each.

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Given: C₁ = 4 μF; C₂ = 12 μF; V = 100 V

Use: Same charge on both; V ∝ 1/C

Ceq = (4 × 12) / 16 = 3 μF; Q = 3 × 10⁻⁶ × 100 = 3 × 10⁻⁴ C

V₁ = Q/C₁ = 3 × 10⁻⁴ / 4 × 10⁻⁶ = 75 V; V₂ = 3 × 10⁻⁴ / 12 × 10⁻⁶ = 25 V

Answer: 75 V across 4 μF and 25 V across 12 μF

2.5Energy of a capacitorStored energy and loss of energy · 6 problems

Q28Energy stored

A 10 μF capacitor is charged to 100 V. How much energy is stored?

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Given: C = 10 × 10⁻⁶ F; V = 100 V

Use: U = ½ C V²

U = ½ × 10 × 10⁻⁶ × (100)² = 5 × 10⁻² J

Answer: 0.05 J

Q29Energy from charge

A capacitor of 5 μF carries a charge of 200 μC. Find the energy stored.

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Given: C = 5 μF; Q = 200 μC

Use: U = Q² / 2C

U = (2 × 10⁻⁴)² / (2 × 5 × 10⁻⁶) = 4 × 10⁻⁸ / 10⁻⁵ = 4 × 10⁻³ J

Answer: 4 × 10⁻³ J

Q30Energy density

Find the energy density of the field in a region where E = 3 × 10⁶ V/m. (ε₀ = 8.85 × 10⁻¹² F/m)

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Given: E = 3 × 10⁶ V/m

Use: u = ½ ε₀ E²

u = ½ × 8.85 × 10⁻¹² × (3 × 10⁶)² = 0.5 × 8.85 × 10⁻¹² × 9 × 10¹² = 39.8 J/m³

Answer: ≈ 40 J/m³

Q31Loss of energy on sharing

A 4 μF capacitor charged to 100 V is connected to an uncharged 4 μF capacitor. Find the energy lost.

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Given: C = 4 μF each; V₁ = 100 V

Use: Charge is conserved; compare energies before and after

Before: U₁ = ½ × 4 × 10⁻⁶ × 100² = 0.02 J

Charge Q = 4 × 10⁻⁶ × 100 = 4 × 10⁻⁴ C is shared by 8 μF, so V = Q/C = 50 V

After: U₂ = ½ × 8 × 10⁻⁶ × 50² = 0.01 J

Energy lost = 0.02 − 0.01 = 0.01 J (half of it)

Answer: 0.01 J

Q32Inserting a dielectric (battery connected)

A 10 μF air capacitor is connected to a 100 V battery. A dielectric of constant 3 is inserted while the battery stays connected. Find the energy before and after.

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Given: C₀ = 10 μF; V = 100 V; K = 3

Use: V stays constant, so U = ½CV² grows with C

Before: U₀ = ½ × 10 × 10⁻⁶ × 100² = 0.05 J

After: C = 30 μF, so U = ½ × 30 × 10⁻⁶ × 100² = 0.15 J (three times)

Answer: 0.05 J before, 0.15 J after

Q33Inserting a dielectric (battery disconnected)

The capacitor of the previous problem is charged to 100 V and the battery is removed. A dielectric of constant 3 is then inserted. Find the new voltage and energy.

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Given: C₀ = 10 μF; V₀ = 100 V; K = 3

Use: Q stays constant: V = Q/C, U = Q²/2C

Q = 10 × 10⁻⁶ × 100 = 10⁻³ C; new C = 30 μF

V = Q/C = 10⁻³ / 30 × 10⁻⁶ = 33.3 V

U = Q²/2C = 10⁻⁶ / (2 × 30 × 10⁻⁶) = 0.0167 J (one third of 0.05 J)

Answer: V = 33.3 V; U = 0.017 J

Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.

2.1Exercises 2.1 to 2.11Potential, conductors, capacitors and energy · 11 questions

Q12.1 Two charges 5 × 10⁻⁸ C and −3 × 10⁻⁸ C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

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Let the point be at a distance x from the 5 × 10⁻⁸ C charge. The potential is zero where V = k(5 × 10⁻⁸ / x − 3 × 10⁻⁸ / r₂) = 0.

Between the charges (r₂ = 0.16 − x): 5 / x = 3 / (0.16 − x), so 5(0.16 − x) = 3x, 0.8 = 8x and x = 0.1 m = 10 cm from the positive charge.

Outside the charges, on the side of the smaller (negative) charge (r₂ = x − 0.16): 5 / x = 3 / (x − 0.16), so 5x − 0.8 = 3x and x = 0.4 m = 40 cm from the positive charge.

There is no such point on the other side of the positive charge, since there the larger charge is closer.

Q22.2 A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.

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In a regular hexagon each vertex is at a distance equal to the side from the centre, r = 0.1 m.

V = 6 × kq / r = 6 × (9 × 10⁹ × 5 × 10⁻⁶) / 0.1 = 2.7 × 10⁶ V.

Q32.3 Two charges 2 μC and −2 μC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?

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(a) The plane normal to AB and passing through its mid-point. Every point on it is equidistant from the two equal and opposite charges, so the potentials cancel and V = 0 on the whole plane.

(b) The field is perpendicular to an equipotential surface, so it is normal to this plane: it is parallel to the line AB, directed from the positive charge (A) towards the negative charge (B).

Q42.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10⁻⁷ C distributed uniformly on its surface. What is the electric field (a) inside the sphere, (b) just outside the sphere, (c) at a point 18 cm from the centre of the sphere?

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(a) The field inside a conductor is zero.

(b) Just outside: E = kq / R² = (9 × 10⁹ × 1.6 × 10⁻⁷) / (0.12)² = 1440 / 0.0144 = 1 × 10⁵ N/C, radially outward.

Q52.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

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C = ε₀A/d = 8 pF. New capacitance C′ = K ε₀ A / (d/2) = 2K × (ε₀A/d) = 2 × 6 × 8 pF = 96 pF.

Q62.6 Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

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(a) 1/C = 1/9 + 1/9 + 1/9 = 3/9, so C = 3 pF.

(b) The same charge is on each: Q = CV = 3 × 10⁻¹² × 120 = 3.6 × 10⁻¹⁰ C. Each capacitor has V = Q/C = 3.6 × 10⁻¹⁰ / 9 × 10⁻¹² = 40 V (equal division: 120 / 3 = 40 V).

Q72.7 Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

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(a) C = 2 + 3 + 4 = 9 pF.

(b) Each has 100 V across it: Q₁ = 2 × 10⁻¹² × 100 = 2 × 10⁻¹⁰ C; Q₂ = 3 × 10⁻¹² × 100 = 3 × 10⁻¹⁰ C; Q₃ = 4 × 10⁻¹² × 100 = 4 × 10⁻¹⁰ C.

Q82.8 In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10⁻³ m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

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C = ε₀A/d = (8.85 × 10⁻¹² × 6 × 10⁻³) / (3 × 10⁻³) = 1.77 × 10⁻¹¹ F = 17.7 pF.

Q = CV = 1.77 × 10⁻¹¹ × 100 = 1.77 × 10⁻⁹ C (about 1.8 × 10⁻⁹ C) on each plate.

Q92.9 Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected, (b) after the supply was disconnected.

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The sheet fills the whole gap, so the new capacitance is C′ = K C = 6 × 17.7 pF = 106 pF.

(a) Supply connected: the voltage stays at 100 V. The charge rises to Q′ = C′V = 1.06 × 10⁻¹⁰ × 100 = 1.06 × 10⁻⁸ C (six times as large). The field in the dielectric stays V/d = 100 / 3 × 10⁻³ = 3.3 × 10⁴ V/m.

(b) Supply disconnected: the charge stays at 1.77 × 10⁻⁹ C. The voltage falls to V′ = Q/C′ = 1.77 × 10⁻⁹ / 1.06 × 10⁻¹⁰ = 16.7 V (one-sixth). The field inside falls to E₀/K = 3.3 × 10⁴ / 6 = 5.6 × 10³ V/m.

Q102.10 A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?

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U = ½ C V² = ½ × 12 × 10⁻¹² × (50)² = ½ × 12 × 10⁻¹² × 2500 = 1.5 × 10⁻⁸ J.

Q112.11 A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

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Initial energy: U₁ = ½ × 600 × 10⁻¹² × (200)² = 1.2 × 10⁻⁵ J.

After connecting, the charge Q = 600 pF × 200 V = 1.2 × 10⁻⁷ C is shared by the total capacitance 1200 pF, so V = Q/C = 100 V. Final energy: U₂ = ½ × 1200 × 10⁻¹² × (100)² = 6 × 10⁻⁶ J.

Energy lost = 1.2 × 10⁻⁵ − 6 × 10⁻⁶ = 6 × 10⁻⁶ J, half of the initial energy. It goes as heat and radiation in the connecting wires.