IIT-JEE 2008 Multiple Choice Questions on Electrostatics

The following two questions from Electrostatics were included under Straight Objective Type (Multiple choice, single answer type) in the IIT-JEE 2008 question paper:

(1) Consider a system of three charges q/3, q/3 and –2q/3 placed at points A, B and C respectively as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB = 60º

(a) The electric field at point O is q/8πε₀R²

(b) The potential energy of the system is zero

(c) The magnitude of the force between the charges at C and B is q²/54πε₀R²

(d) The potential at point O is q/12πε₀R

The electric field at O is due to the negative charge at C only since the equal positive charges situated at A and B will produce equal and opposite fields at O (and they will mutually cancel). The field at O is therefore negative and the option (a) is obviously wrong. Option (b) also is obviously wrong.

The magnitude of the force (F) between the charges at C and B is given by

F = (1/4πε₀) [(2q/3)(q/3)/(2Rsin60º)²].

Thus F = q²/54πε₀R² as given in option (c).

(2) A parallel plate capacitor with plates of unit area and separation d is filled with a liquid of dielectric constant K = 2. The level of liquid is d/3 initially. Suppose the liquid level decreases at a constant speed V. The time constant as a function of time t is

(a) 6ε₀R/(5d + 3Vt)

(b) (15d + 9Vt) ε₀R/(2d² – 3dVt – 9V²t²)

(c) 6ε₀R/(5d – 3Vt)

(d) (15d – 9Vt) ε₀R/(2d² + 3dVt – 9V²t²)

When a dielectric slab of thickness t₁ and dielectric constant K is introduced in between the plates of a parallel plate air capacitor of plate area A and plate separation d, the effective capacitance changes from ε₀A/d to ε₀A/[d–t₁ + (t₁/K)]

The effective capacitance C of a parallel plate capacitor with parallel dielectric slabs of thickness t₁, t₂, t₃ etc. of dielectric constants K₁, K₂, K₃ etc. respectively is given by the series combined value of capacitors with these dielectrics. Therefore,

1/C = 1/(K₁ε₀A/t₁) + 1/(K₂ε₀A/t₂) + 1/(K₃ε₀A/t₃) + etc.

With a single slab of thickness t₁ and dielectric constant K introduced between the plates, we have

1/C = 1/(Kε₀A/t₁) + 1/[ε₀A/(d – t₁)] from which C = ε₀A/[d–t₁ + (t₁/K)]

In the present problem, the thickness of the layer of liquid (which serves as the dielectric slab) at the instant t is (d/3) – V t and the thickness of the air space is (2d/3) + V t.

The effective capacitance is therefore given by

C = ε₀A/[{(2d/3) + Vt} + {(d/3) – Vt}/k)]

Substituting for A = 1 and K = 2, we obtain C = 6ε₀/(5d + 3Vt).

The time constant of the circuit = RC = 6ε₀R/(5d + 3Vt).

Reality is merely an illusion, albeit a very persistent one. 

–Albert Einstein 

Multiple Choice Questions (MCQ) involving Inductance

(1) A long straight solenoid of cross section area10 cm² has 10 turns per cm. A short 50 turn coaxial coil of cross section area 2 cm² is fixed inside the solenoid at the middle. The mutual inductance of the solenoid and the coil is (in micro henry)

(a) 400 (b) 40 (c) 4π (d) π (e) ) 0.4π

Mutual inductance between an infinitely long straight solenoid and a short secondary coil placed coaxially inside at the middle is given by

M = μ₀nNA where μ₀ is the magnetic permeability of free space (or air), n is the number of turns per metre of the solenoid, N is the total number of turns in the secondary coil and A is the cross section area of the secondary coil. (If the secondary coil is outside the solenoid, the cross section area of the solenoid will appear in place of the area of the secondary coil).

[Note that the mutual inductance as given by the above expression is the magnetic flux (linked with the secondary coil) per unit current in the solenoid: M = NBA where B is the magnetic field (μ₀n×1) produced by unit current in the solenoid]

Substituting for the known quantities in the expression for mutual inductance, we have

M = 4π×10^–7×1000 ×50 ×(2×10^–4) = 4π×10^–6 H = 4π μH.

(2) The secondary coil in the above question is moved with uniform velocity coaxially over a distance of 5 cm in 2 seconds when a current of 2 ampere flows through the solenoid. The emf induced in the coil during the motion is

(a) 200π μV (b) 100π μV (c) 20π μV (d) 10π μV (e) zero

No emf will be induced in the coil since the same flux is linked with the coil throughout its motion and there is no flux change.

(3) In question No.1, suppose the solenoid carries a current of 4 A. If this current is switched off in 100 ms, the emf induced in the secondary coil will be

(a) 0.2π μV (b) 0.4π μV (c) π mV (d) 0.08π mV (e) 0.16π mV

We have ε = –M(dI/dt) = = 4π×10^–6 [(4 – 0)/(100×10^–3)] = 160π×10^–6 volt = 0.16π mV.

4. A battery of emf V volt is connected in series with a coil of inductance L and resistance R at the instant t = 0. The current in the circuit when t = 2τ where τ is the time constant of the circuit is (base of natural logarithm = e)

(a) (V/R)[1– (1/e²)]

(b) (V/R (1/e²)

(c) (V/R)[1– (1/e)]

(d) (V/R)(1/e)]

(e) 2e(V/R)

The expression for the growth of current (with time) in an LR circuit is

I = I₀[1– e^–Rt/L]

where I₀ is the final steady (maximum possible) current which is equal to V/R.

The time constant of the LR circuit is L/R. After 2 time constants, the current will be

I = I₀[1– e^{–R× 2(L/R)/L}] = (V/R) [1– e^{– 2}] = (V/R)[1– (1/e²)]

You will find more questions (with solution) at AP Physics Resources: AP Physics C - Multiple Choice Questions on Inductance