AIPMT 2009 - Multiple Choice Questions from Electrostatics

Three questions were included from electrostatics in the All India Pre-Medical/Pre-Dental 2009 Entrance Examination (Preliminary). They are given below with solution. The first question will appear to be a rather difficult and time consuming one. Those who are preparing for AP Physics Exam may make a special note of this question.
(1) Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ, − σ and σ respectively. If V_A, V_B and V_C denote the potentials of the three shells, then for c = a + b, we have

(1) V_C = V_B ≠ V_A

(2) V_C ≠ V_B ≠ V_A

(3) V_C = V_B = V_A

(4) V_C = V_A ≠ V_B

The potential V of a spherical shell of radius r having surface charge density σ is given by

V = (1/4πε₀)(4πr²σ/r) = σr/ε₀ where ε₀ is the permittivity of free space.

Potential V_A of the shell A is given by

V_A = σa/ε₀ – σb/ε₀ + σc/ε₀ = (σ/ε₀)[c – (b – a)]

Potential V_B of the shell B is given by

V_B = – σb/ε₀ + (1/4πε₀)(4πa²σ/b) + σc/ε₀

Or, V_B = σ/ε₀ [c – (b² – a²)/b]

Potential V_C of the shell C is given by

V_C = σc/ε₀ – (1/4πε₀)(4πb²σ/c) + (1/4πε₀)(4πa²σ/c)

Or, V_C = σ/ε₀ [c – (b² – a²)/c] = σ/ε₀ [c – (b + a)(b– a)/c]

Since c = a + b we obtain

V_C = σ/ε₀ [c – (b– a)]

The correct option therefore is V_C = V_A ≠ V_B.

(2) Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be

(1) 3 C, V/3

(2) C/3, 3 V

(3) 3 C, 3V

(4) C/3, V/3

This is a very simple question. The effective calacitance C_eff of the combination of three capacitors in series is given by the reciprocal relation,

1/C_eff = 1/C₁ +1/C₂ +1/C₃

Here C₁ = C₂ = C₃ = C so that C_eff = C/3

The breakdown voltage V_eff for the series combination is the sum of the individual breakdown voltages:

V_eff = 3V

(3) The electric potential at a point (x, y, z) is given by V = − x²y − xz³ + 4. The electric field E at that point is:

(1) E = i 2xy + j (x² + y²) + k (3xz − y²)

(2) E = i z ³ + j xyz + k z²

(3) E = i (2xy − z³) + j xy² + k 3z²x

(4) E = i (2xy + z³) + j x² + k 3xz²

The electric field E is the negative gradient of potential:

E = − ∂V/∂r = − (i ∂V/∂x + j ∂V/∂y + k ∂V/∂z)

This gives E = i (2xy + z³) + j x² + k 3xz² as given in option (4).

You will find similar useful multiple choice questions (with solution) in electrostatics at aphysicsresources and at physicsplus