NEET Physics: Electrostatics and Current Electricity · Chapter 2

Electrostatic Potential and Capacitance

What to remember

Potential at distance r from a point charge Q is V = kQ/r. It is a scalar, so the potential of many charges is the ordinary algebraic sum.
The field is the negative gradient of potential: E = –dV/dr. The field points from high to low potential and is perpendicular to equipotential surfaces.
Parallel-plate capacitor: C = ε₀A/d in vacuum and C = Kε₀A/d with a full dielectric. Series: 1/C = Σ(1/Cᵢ). Parallel: C = ΣCᵢ.
Energy stored U = ½CV² = Q²/2C = ½QV. Energy density in the field = ½ε₀E². When capacitors share charge, energy is lost as heat.

Electrostatic potential and potential difference

The electrostatic potential difference between points A and B is the work done by an external agent in moving a unit positive test charge from A to B without acceleration:

V_B – V_A = W_AB/q₀

The unit is the volt (V) = J C⁻¹. The potential at a point is the work done in bringing a unit positive charge from infinity to that point. Potential is a scalar. The work done by the electric field between two points depends only on the end points, not on the path, because the electrostatic force is conservative.

Work done in moving charge q through a potential difference ΔV is W = qΔV. One electron-volt (eV) is the energy gained by an electron through 1 V: 1 eV = 1.6 × 10⁻¹⁹ J.

Potential due to a point charge, a dipole and a system of charges

Source	Potential	Notes
Point charge Q	V = kQ/r	Positive Q gives positive V; falls as 1/r
System of charges q₁, q₂ ...	V = k(q₁/r₁ + q₂/r₂ + ...)	Algebraic sum, signs included
Dipole, axial point (r much greater than a)	V = kp/r²	Positive on the side of +q
Dipole, equatorial line	V = 0	Because the potentials of +q and –q cancel
General point at angle θ from p	V = kp cos θ/r²	Falls as 1/r²
Charged conducting sphere, radius R, charge Q	Outside and on the surface: kQ/r; inside: kQ/R (constant)	Whole conductor is at one potential

Compare: dipole field falls as 1/r³ and dipole potential falls as 1/r².

Equipotential surfaces

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 an equipotential surface.
Two equipotential surfaces never intersect.
For a point charge they are concentric spheres. For a uniform field they are planes perpendicular to the field.
The field points from higher potential to lower potential. Surfaces are closer together where the field is stronger.

Relation between field and potential

The field is the negative rate of change of potential with distance:

E = –dV/dr

In a uniform field, the magnitude is E = ΔV/d, where d is the separation along the field. The negative sign shows that E points in the direction in which V decreases. From this relation 1 V m⁻¹ = 1 N C⁻¹. If V is constant over a region, E is zero there.

Potential energy of charges and of a dipole

Two point charges q₁ and q₂ at separation r: U = kq₁q₂/r. It is positive for like charges (work is needed to bring them closer) and negative for unlike charges.

Three or more charges: add the energy of every pair, U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃).

Dipole in a uniform field: U = –pE cos θ. The work done in rotating the dipole from θ₁ to θ₂ is W = pE(cos θ₁ – cos θ₂).

Single charge q at a point of potential V: U = qV.

Conductors and electrostatic shielding

In electrostatics (all charges at rest):

1. The electric field inside a conductor is zero.
2. At the surface the field is normal to the surface, with magnitude σ/ε₀.
3. Any excess charge resides only on the outer surface.
4. The potential is constant throughout the conductor, including the surface.
5. The field inside a cavity of a conductor is zero when the cavity has no charge in it.

Electrostatic shielding: a hollow conducting shell protects the space inside from external electric fields. Free charges rearrange themselves on the shell until the field inside is cancelled. This is why a sensitive instrument is kept inside a metal box and why a car or aircraft body protects people inside during lightning.

Dielectrics and polarisation

A dielectric is an insulating material that has no free charge carriers. In an external field, dielectrics become polarised:

Non-polar molecules (for example, N₂, O₂, CO₂): centres of positive and negative charge coincide with no field. The field creates an induced dipole moment.
Polar molecules (for example, H₂O, HCl): permanent dipoles that are randomly oriented. The field tends to align them.

In both cases the dielectric develops a net dipole moment per unit volume called polarisation. Charge appears on its two faces (bound charge). It produces a field inside the dielectric opposite to the applied field, so the net field is reduced:

E = E₀/K

The dielectric constant K = ε/ε₀ is the ratio of the field in vacuum to the field in the dielectric for the same charge. K = 1 for vacuum and K is greater than 1 for any dielectric (about 80 for water). For a conductor, K is taken as infinite.

Capacitors and capacitance

A capacitor is two conductors separated by an insulator. The charge on the plates is +Q and –Q. The capacitance is C = Q/V. Its unit is the farad (F) = C V⁻¹. Practical values are in μF, nF and pF.

Parallel-plate capacitor (plate area A, separation d):

Case	Capacitance
Air or vacuum	C₀ = ε₀A/d
Completely filled with dielectric of constant K	C = Kε₀A/d = KC₀
Dielectric slab of thickness t (less than d) inserted	C = ε₀A/[d – t(1 – 1/K)]
Conducting slab of thickness t inserted (K = infinite)	C = ε₀A/(d – t)

The field between the plates is E = σ/ε₀ = Q/(ε₀A) in air. Capacitance depends only on the geometry and the medium. It does not depend on Q or V.

Effect of a slab with the battery connected (V constant): C rises by a factor K, charge Q = CV rises by a factor K, field E = V/d is unchanged, and stored energy rises by K.

Effect of a slab after the battery is removed (Q constant): C rises by K, voltage falls to V/K, field falls to E/K, and stored energy falls to U/K. The slab is pulled into the capacitor.

Combination of capacitors

Arrangement	Rule	Special feature
Series	1/C = 1/C₁ + 1/C₂ + ...	Same charge on each; voltage divides; C is less than the smallest
Parallel	C = C₁ + C₂ + ...	Same voltage on each; charge divides; C is greater than the largest

In series two capacitors give C = C₁C₂/(C₁ + C₂). In series, the smaller capacitor has the larger voltage. In parallel, the larger capacitor has the larger charge. For n identical capacitors of value C: series gives C/n and parallel gives nC.

Energy stored in a capacitor and energy density

The work done in charging a capacitor is stored as energy in the electric field:

U = ½CV² = Q²/(2C) = ½QV

Use the form that keeps the fixed quantity visible: use Q²/2C when Q is fixed and ½CV² when V is fixed.

The energy per unit volume in the field (energy density) is u = ½ε₀E². In a dielectric this is ½Kε₀E². It is the same everywhere inside a parallel-plate capacitor, ignoring edge effects. The unit is J m⁻³.

Common-potential and charge sharing

When two charged conductors (or capacitors) are joined with a wire, charge flows until both are at the same potential V_c.

V_c = (C₁V₁ + C₂V₂)/(C₁ + C₂) = total charge/total capacitance

Charge is conserved, but energy is not: the loss of energy appears as heat and radiation. The loss is

ΔU = ½ [C₁C₂/(C₁ + C₂)] (V₁ – V₂)²

This loss is always positive unless V₁ = V₂. If two capacitors are joined plate to plate with opposite charges, use signed charges, i.e. subtract in the total.

Formula summary

Quantity	Formula
Potential of a point charge	V = kQ/r
Potential energy of two charges	U = kq₁q₂/r
Field from potential	E = –dV/dr
Dipole axial potential	V = kp/r²
Capacitance	C = Q/V
Parallel plate	C = ε₀A/d; with dielectric C = Kε₀A/d
Series	1/C = Σ 1/Cᵢ
Parallel	C = Σ Cᵢ
Energy	U = ½CV² = Q²/2C
Energy density	u = ½ε₀E²
Common potential	V_c = (C₁V₁ + C₂V₂)/(C₁ + C₂)

Worked examples

Example 1 (potential). Find the potential at 30 cm from a point charge of 5 μC.

V = 9 × 10⁹ × 5 × 10⁻⁶/0.3 = 1.5 × 10⁵ V.

Example 2 (potential energy). Charges 2 μC and 3 μC are 6 cm apart in vacuum. U = 9 × 10⁹ × 2 × 10⁻⁶ × 3 × 10⁻⁶/0.06 = 0.9 J.

Example 3 (parallel plate). Area 0.01 m², separation 1 mm, vacuum. C = 8.85 × 10⁻¹² × 0.01/10⁻³ = 8.85 × 10⁻¹¹ F = 88.5 pF.

Example 4 (charge sharing). A 2 μF capacitor charged to 100 V is joined to an uncharged 3 μF capacitor. Total charge is 200 μC and total capacitance is 5 μF, so V_c = 40 V. Initial energy = ½ × 2 × 10⁻⁶ × 10⁴ = 10⁻² J. Final energy = ½ × 5 × 10⁻⁶ × 1600 = 4 × 10⁻³ J. Energy lost = 6 × 10⁻³ J. Check with the formula: ½ × (6/5) × 10⁻⁶ × 10⁴ = 6 × 10⁻³ J.

Example 5 (series and energy). A 3 μF and a 6 μF capacitor are in series across 90 V. C = 18/9 = 2 μF. Charge = 2 × 10⁻⁶ × 90 = 180 μC on each. Voltage across the 3 μF capacitor = 180/3 = 60 V and across the 6 μF capacitor = 30 V. Total energy = ½ × 2 × 10⁻⁶ × 8100 = 8.1 × 10⁻³ J.

NEET traps

1. Potential is a scalar; field is a vector. Add potentials as numbers, with signs; add fields as vectors.
2. Zero field does not mean zero potential, and zero potential does not mean zero field. On the equatorial line of a dipole V = 0 but E is not zero. Inside a charged conductor E = 0 but V is not zero.
3. Dipole potential falls as 1/r² and field as 1/r³.
4. Slab inserted with battery on versus off. With the battery on, V is constant; with it off, Q is constant. Energy rises in the first case and falls in the second.
5. Series capacitors carry the same charge; parallel capacitors have the same voltage. The rule for capacitors is opposite to that for resistors in the formula for C.
6. Energy lost in charge sharing is not destroyed. It becomes heat and radiation. Charge is always conserved.
7. Potential energy of two like charges is positive, and of two unlike charges is negative.
8. Capacitance does not depend on the charge or the voltage, only on geometry and the medium.
9. Excess charge on a conductor lies on the outer surface, and the field is zero inside even though the potential is not.
10. Equipotential surfaces and field lines are perpendicular, and surfaces never intersect.

One-liners

1. The SI unit of potential is the volt, which equals J C⁻¹.
2. Potential difference of 1 V does 1 J of work per coulomb.
3. 1 eV = 1.6 × 10⁻¹⁹ J.
4. The potential of a point charge falls as 1/r.
5. The equatorial line of a dipole is at zero potential.
6. Electric field is the negative gradient of the potential.
7. Work done along an equipotential surface is zero.
8. The electrostatic potential inside a charged conducting sphere is constant.
9. The field at the surface of a charged conductor is σ/ε₀.
10. The unit of capacitance is the farad, which equals C V⁻¹.
11. The capacitance of a parallel-plate capacitor is inversely proportional to the plate separation.
12. A dielectric increases capacitance by the factor K.
13. Capacitors in parallel add their capacitances.
14. The energy stored in a capacitor is ½CV².
15. The energy density of an electric field is ½ε₀E².

Practice questions

The electric potential at a distance of 0.3 m from a point charge of 5 μC in vacuum is (k = 9 × 10⁹ N m² C⁻²)
1. 5 × 10⁵ V
2. 4.5 × 10⁴ V
3. 1.5 × 10⁶ V
4. 1.5 × 10⁵ V
Answer
D. 1.5 × 10⁵ V
V = kq/r = 9 × 10⁹ × 5 × 10⁻⁶/0.3 = 1.5 × 10⁵ V. (5 × 10⁵ would be the field kq/r² in N C⁻¹.)
The work done by an external agent in moving a charge of 2 μC slowly through a potential difference of 500 V is
1. 1 × 10⁻⁹ J
2. 1 × 10⁻³ J
3. 2.5 × 10⁸ J
4. 1 × 10⁻⁶ J
Answer
B. 1 × 10⁻³ J
W = qΔV = 2 × 10⁻⁶ × 500 = 1 × 10⁻³ J.
An electron starting from rest is accelerated through a potential difference of 200 V. Its kinetic energy gain is (e = 1.6 × 10⁻¹⁹ C)
1. 3.2 × 10⁻¹⁷ J
2. 3.2 × 10⁻¹⁹ J
3. 200 J
4. 1.6 × 10⁻¹⁷ J
Answer
A. 3.2 × 10⁻¹⁷ J
K = eV = 1.6 × 10⁻¹⁹ × 200 = 3.2 × 10⁻¹⁷ J (200 eV).
The electrostatic potential energy of a system of two point charges 2 μC and 3 μC placed 6 cm apart in vacuum is (k = 9 × 10⁹ N m² C⁻²)
1. –0.9 J
2. 9 J
3. 0.9 J
4. 0.09 J
Answer
C. 0.9 J
U = kq₁q₂/r = 9 × 10⁹ × 6 × 10⁻¹²/0.06 = 0.9 J, positive for like charges.
The potential difference between two parallel plane equipotential surfaces 2 cm apart is 40 V. The magnitude of the uniform electric field between them is
1. 20 V m⁻¹
2. 0.8 V m⁻¹
3. 2 × 10⁵ V m⁻¹
4. 2000 V m⁻¹
Answer
D. 2000 V m⁻¹
E = ΔV/d = 40/0.02 = 2000 V m⁻¹.
A parallel-plate air capacitor has plate area 0.02 m² and plate separation 2 mm. With ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻², its capacitance is
1. 8.85 × 10⁻¹⁰ F
2. 8.85 × 10⁻⁸ F
3. 8.85 × 10⁻¹¹ F
4. 1.77 × 10⁻¹⁰ F
Answer
C. 8.85 × 10⁻¹¹ F
C = ε₀A/d = 8.85 × 10⁻¹² × 0.02/(2 × 10⁻³) = 8.85 × 10⁻¹¹ F.
Two capacitors of 6 μF and 3 μF are connected in series. The equivalent capacitance is
1. 4.5 μF
2. 2 μF
3. 0.5 μF
4. 9 μF
Answer
B. 2 μF
C = (6 × 3)/(6 + 3) = 2 μF. Series gives a value less than the smaller capacitor.
The energy stored in a 10 μF capacitor charged to 200 V is
1. 0.2 J
2. 0.4 J
3. 0.02 J
4. 2 J
Answer
A. 0.2 J
U = ½CV² = ½ × 10 × 10⁻⁶ × 4 × 10⁴ = 0.2 J.
The energy density in a region of uniform electric field 10⁴ V m⁻¹ in vacuum is (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)
1. 4.4 × 10⁻⁴ J m⁻³
2. 4.4 × 10⁻⁸ J m⁻³
3. 8.85 × 10⁻⁴ J m⁻³
4. 8.85 × 10⁻⁸ J m⁻³
Answer
A. 4.4 × 10⁻⁴ J m⁻³
u = ½ε₀E² = 0.5 × 8.85 × 10⁻¹² × 10⁸ = 4.4 × 10⁻⁴ J m⁻³.
A parallel-plate capacitor is completely filled with a dielectric of constant 4 and its plate separation is also halved. The capacitance becomes
1. 4 times
2. 2 times
3. 8 times
4. 16 times
Answer
C. 8 times
C = Kε₀A/d. K ×4 and d ÷2 give C × 8.
The electric potential at any point on the equatorial line of an electric dipole is
1. kp/r
2. kp/r²
3. 2kp/r²
4. zero
Answer
D. zero
The potentials of +q and –q at equal distances cancel, so V = 0.
The work done in moving a charge between two points on the same equipotential surface is
1. negative
2. zero
3. dependent on the path
4. positive and equal to qV
Answer
B. zero
No potential difference means W = qΔV = 0.
A charged conducting sphere of radius 0.1 m carries 2 μC. The potential at its centre is (k = 9 × 10⁹ N m² C⁻²)
1. 1.8 × 10⁵ V
2. 9 × 10⁴ V
3. 1.8 × 10⁶ V
4. zero
Answer
A. 1.8 × 10⁵ V
Inside a conductor V equals the surface value kQ/R = 9 × 10⁹ × 2 × 10⁻⁶/0.1 = 1.8 × 10⁵ V.
A 4 μF capacitor charged to 50 V is connected across an uncharged 6 μF capacitor. The common potential is
1. 25 V
2. 20 V
3. 50 V
4. 30 V
Answer
B. 20 V
V = C₁V₁/(C₁ + C₂) = (4 × 50)/10 = 20 V.
One farad is equal to
1. one coulomb per metre
2. one joule per coulomb
3. one volt per coulomb
4. one coulomb per volt
Answer
D. one coulomb per volt
C = Q/V, so 1 F = 1 C V⁻¹.
The electric field just outside the surface of a charged conductor with surface charge density 8.85 × 10⁻⁹ C m⁻² is (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)
1. 5 × 10² N C⁻¹
2. 1 × 10⁶ N C⁻¹
3. 1 × 10³ N C⁻¹
4. 2 × 10³ N C⁻¹
Answer
C. 1 × 10³ N C⁻¹
E = σ/ε₀ = 8.85 × 10⁻⁹/8.85 × 10⁻¹² = 10³ N C⁻¹ (σ/2ε₀ applies to an isolated sheet, not a conductor surface).
Given below are two statements: Statement I: Electric potential is a vector quantity. Statement II: The potential at a point due to several charges is the algebraic sum of the potentials due to the individual charges. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Potential is a scalar, so I is wrong. II is correct.
Given below are two statements: Statement I: The electric field inside a conductor in electrostatics is zero. Statement II: The potential inside a charged conductor is zero. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
E = 0 inside, so V is constant, equal to the surface value, and not zero.
Given below are two statements: Statement I: Electric field lines point from lower to higher potential. Statement II: Electric field is perpendicular to equipotential surfaces. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
E points in the direction of decreasing potential, so I is wrong. II is correct.
Given below are two statements: Statement I: With the battery kept connected, inserting a dielectric slab between the plates increases the charge on the plates. Statement II: With the battery kept connected, the potential difference across the plates stays the same. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
V is fixed by the battery. C rises, so Q = CV rises.
Which of the following statements are correct? 1. The capacitance of a parallel-plate capacitor depends on the charge on it. 2. It depends on the plate area. 3. It depends on the medium between the plates.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
B. 2 and 3 only
C = Kε₀A/d depends on A, d and the medium, not on Q or V.
Given below are two statements: Statement I: When two charged capacitors are joined, the total charge is conserved. Statement II: When two charged capacitors at different potentials are joined, the total energy is conserved. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Charge is conserved but some energy is lost as heat and radiation.
Which of the following statements are correct? 1. Capacitors in series carry the same charge. 2. Capacitors in parallel have the same potential difference. 3. The equivalent capacitance in series is greater than the largest capacitance.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 2 only
Statements 1 and 2 are true. In series the equivalent capacitance is smaller than the smallest one.
Given below are two statements: Statement I: Non-polar molecules have a permanent dipole moment. Statement II: A dielectric increases the electric field inside it by a factor K. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Non-polar molecules have only induced dipoles. A dielectric reduces the field to E₀/K.
Which of the following statements are correct? 1. A dielectric reduces the electric field inside it. 2. The dielectric constant of a conductor is zero. 3. The dielectric constant of vacuum is 1.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
Polarisation reduces the field. A conductor is treated as K = infinite, not zero.
Given below are two statements: Statement I: When a dielectric slab is fully inserted after the battery is disconnected, the potential difference across the plates rises. Statement II: With the battery disconnected, the charge on the plates remains constant. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Q constant and C increases means V = Q/C falls, so I is wrong. II is correct.
Assertion (A): The electric potential at every point on the equatorial line of a dipole is zero. Reason (R): The potentials due to +q and –q at equal distances cancel each other.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
V = kq/r – kq/r = 0 on the equatorial line.
Assertion (A): The capacitance of a parallel-plate capacitor is ε₀A/d in air. Reason (R): The capacitance of a capacitor depends on the charge given to its plates.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
A is true. R is false because C depends only on geometry and the medium.
Assertion (A): No work is done in moving a charge along an equipotential surface. Reason (R): The electrostatic force is a conservative force.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both true, but the work is zero because E is perpendicular to the surface (no potential difference), which is not the same as conservativeness.
Assertion (A): The excess charge on a conductor stays inside its volume. Reason (R): The electric field inside a conductor in electrostatics is zero.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
Excess charge sits on the outer surface, so A is false. R is true.
Assertion (A): Energy is lost when two capacitors at different potentials are connected by a wire. Reason (R): Charge is conserved in the process.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both true. Conservation of charge fixes the common potential; it does not explain the energy loss, which goes to heat and radiation.
Assertion (A): The potential difference between any two points of a conductor in electrostatics is zero. Reason (R): The electric field inside a conductor is zero.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
E = –dV/dr = 0 means V is constant everywhere in the conductor.
Match List I with List II. List I: a. Potential of a point charge b. Potential of a dipole on its axis c. Potential of a dipole on its equatorial line d. Potential energy of two point charges. List II: i. kp/r² ii. kq/r iii. zero iv. kq₁q₂/r.
1. a-i, b-ii, c-iii, d-iv
2. a-ii, b-i, c-iv, d-iii
3. a-iii, b-i, c-ii, d-iv
4. a-ii, b-i, c-iii, d-iv
Answer
D. a-ii, b-i, c-iii, d-iv
Point charge kq/r; axial dipole kp/r²; equatorial zero; two charges kq₁q₂/r.
Match List I with List II. List I: a. Series combination b. Parallel combination c. Energy stored in a capacitor d. Energy density in a field. List II: i. ½ε₀E² ii. ½CV² iii. 1/C = 1/C₁ + 1/C₂ + ... iv. C = C₁ + C₂ + ....
1. a-iii, b-iv, c-ii, d-i
2. a-ii, b-iii, c-iv, d-i
3. a-iii, b-iv, c-i, d-ii
4. a-iv, b-iii, c-ii, d-i
Answer
A. a-iii, b-iv, c-ii, d-i
Series: reciprocal sum. Parallel: sum. Energy ½CV². Density ½ε₀E².
Match List I with List II. List I: a. Capacitance b. Potential difference c. Energy density d. Electric field. List II: i. V m⁻¹ ii. C V⁻¹ iii. J m⁻³ iv. J C⁻¹.
1. a-i, b-iv, c-iii, d-ii
2. a-iv, b-ii, c-iii, d-i
3. a-ii, b-iv, c-iii, d-i
4. a-ii, b-iv, c-i, d-iii
Answer
C. a-ii, b-iv, c-iii, d-i
C = Q/V (C V⁻¹); V = W/q (J C⁻¹); u in J m⁻³; E in V m⁻¹.
Match List I with List II. List I: a. Dielectric slab inserted, battery connected: stored energy b. Dielectric slab inserted, battery disconnected: stored energy c. Conducting slab of thickness d/2 inserted: capacitance d. Plate separation doubled: capacitance. List II: i. halves ii. doubles iii. decreases to 1/K times iv. increases to K times.
1. a-iii, b-iv, c-ii, d-i
2. a-iv, b-iii, c-ii, d-i
3. a-ii, b-iii, c-iv, d-i
4. a-iv, b-iii, c-i, d-ii
Answer
B. a-iv, b-iii, c-ii, d-i
V fixed: U = ½CV² rises K times. Q fixed: U = Q²/2C falls to 1/K. C = ε₀A/(d – d/2) = 2C₀. C ∝ 1/d gives half.
Match List I with List II (k = 9 × 10⁹ N m² C⁻²). List I: a. Potential at 0.9 m from a 1 μC charge b. 3 μF and 6 μF in series c. Energy stored in 4 μF at 100 V d. 3 μF and 6 μF in parallel. List II: i. 0.02 J ii. 10⁴ V iii. 9 μF iv. 2 μF.
1. a-iv, b-ii, c-i, d-iii
2. a-ii, b-iii, c-i, d-iv
3. a-ii, b-iv, c-i, d-iii
4. a-i, b-iv, c-ii, d-iii
Answer
C. a-ii, b-iv, c-i, d-iii
a: 9 × 10⁹ × 10⁻⁶/0.9 = 10⁴ V. b: 18/9 = 2 μF. c: ½ × 4 × 10⁻⁶ × 10⁴ = 0.02 J. d: 9 μF.
Match List I with List II. List I: a. Field inside a conductor b. Field at the surface of a conductor c. Potential inside a conductor d. Excess charge on a conductor. List II: i. Resides on the outer surface ii. Same as the surface potential iii. Zero iv. Normal to the surface, σ/ε₀.
1. a-iv, b-iii, c-ii, d-i
2. a-iii, b-iv, c-ii, d-i
3. a-ii, b-iv, c-iii, d-i
4. a-iii, b-iv, c-i, d-ii
Answer
B. a-iii, b-iv, c-ii, d-i
Inside E = 0; surface E = σ/ε₀ normal; V constant; charge on the outer surface.
Charges +4 μC, –1 μC, +2 μC and –1 μC are at the four corners of a square whose diagonal is 0.6 m. The potential at the centre is (k = 9 × 10⁹ N m² C⁻²)
1. 1.2 × 10⁶ V
2. zero
3. 3.6 × 10⁵ V
4. 1.2 × 10⁵ V
Answer
D. 1.2 × 10⁵ V
Each corner is 0.3 m from the centre. V = k(4 – 1 + 2 – 1) × 10⁻⁶/0.3 = 9 × 10⁹ × 4 × 10⁻⁶/0.3 = 1.2 × 10⁵ V.
A 2 μC charge is released from rest at a point of potential 300 V and moves to a point of potential 100 V. Its kinetic energy there is
1. 4 × 10⁻⁴ J
2. 2 × 10⁻⁴ J
3. 6 × 10⁻⁴ J
4. 8 × 10⁻⁴ J
Answer
A. 4 × 10⁻⁴ J
K = q(V₁ – V₂) = 2 × 10⁻⁶ × 200 = 4 × 10⁻⁴ J.
An air capacitor of 8 pF is charged to 100 V and the battery is removed. A dielectric of constant 4 is then fully inserted. The energy stored is now
1. 1 × 10⁻⁸ J
2. 1.6 × 10⁻⁷ J
3. 2.5 × 10⁻⁹ J
4. 4 × 10⁻⁸ J
Answer
A. 1 × 10⁻⁸ J
Initial U = ½ × 8 × 10⁻¹² × 10⁴ = 4 × 10⁻⁸ J. Q is constant, so U falls to U/K = 1 × 10⁻⁸ J.
A 6 μF capacitor at 50 V and a 3 μF capacitor at 20 V are joined positive plate to positive plate. The common potential is
1. 45 V
2. 35 V
3. 40 V
4. 30 V
Answer
C. 40 V
V = (6 × 50 + 3 × 20)/(6 + 3) = 360/9 = 40 V.
A 4 μF capacitor charged to 100 V is connected to an identical uncharged capacitor. The energy lost as heat is
1. 0.03 J
2. 0.02 J
3. 0.005 J
4. 0.01 J
Answer
D. 0.01 J
Initial U = ½ × 4 × 10⁻⁶ × 10⁴ = 0.02 J. Final: 50 V across 8 μF gives 0.01 J. Loss = 0.01 J.
A 2 μF and a 4 μF capacitor are connected in series across a 12 V battery. The potential difference across the 2 μF capacitor is
1. 6 V
2. 8 V
3. 12 V
4. 4 V
Answer
B. 8 V
C = 4/3 μF, Q = 16 μC. V = Q/C = 16/2 = 8 V across the 2 μF (and 4 V across the 4 μF).
An air capacitor with plates 6 mm apart has capacitance 100 pF. A conducting slab 2 mm thick is placed between the plates without touching them. The capacitance becomes
1. 66.7 pF
2. 300 pF
3. 100 pF
4. 150 pF
Answer
D. 150 pF
C = ε₀A/(d – t) = C₀ d/(d – t) = 100 × 6/4 = 150 pF.

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