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