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

Electrostatics: MCQ on Charged Spherical Drops

Often questions involving the calculation of the potential of a drop obtained by the combination of a number of identical charged spherical droplets are seen in college entrance question papers. If n identical small drops each of radius r carrying charge q coalesce to form a single large drop of radius R, we have

R = n^1/3 r

[You will get this by equating the volumes: n×(4/3)πr³ = (4/3)πR³]

The potential V on the surface of each small drop is given by

V = (1/4πε₀)(q/r)

The total charge on the larger drop is nq. Therefore, the potential V’ on its surface is

V’ = (1/4πε₀)(nq/R) = (1/4πε₀)(nq/n^1/3r) since R = n^1/3 r

Therefore, V’ = n^2/3V

The electric field E on the surface of each small drop is given by

E = (1/4πε₀)(q/r²)

The electric field E’ on the surface of the large drop is given by

E’ = (1/4πε₀)(nq/R²) = (1/4πε₀)(nq/n^2/3r² ) since R = n^1/3 r

Therefore, E’ = n^1/3E

If you can remember the above expressions for the electric potential and field on the larger drop, multiple choice questions involving them can be answered in no time. But it is always more rewarding to remember the basic things so that you can calculate the required quantities in all situations.

The calculation of V’ in the above form itself appeared as a multiple choice question in Kerala Engineering Entrance 2008 question paper.

Now answer the following multiple choice question involving charged drops. You may try them yourself and then check with the given solution. Here are the questions:

(1) Sixty four identical drops each charged by q coulomb to potential V volt coalesce to form a single large drop. The charge and potential on the large drop are respectively

(a) 64q, 64V

(b) 64q, V

(c) 8q, 16V

(d) 16q, 16V

(e) 64q, 16V

The charge on the large drop is the sum of the charges on the small drops and is equal to 64q.

As shown above, the potential on the large drop is V’ = n^2/3V = 64^2/3V =16V [Option (e)].

(2) Two identical soap bubbles A and B are uniformly charged with the same amount of charge. But the charge on A is positive where as the charge on B is negative. (The electrical interaction between the bubbles is negligible). Because of charging, the excess of pressure inside

(a) both bubbles will increase

(b) both bubbles will decrease

(c) both bubbles will remain unchanged

(d) A will increase and that inside B will decrease

(e) B will increase and that inside A will decrease

Because of the repulsive force between like charges, the bubbles will expand and hence the excess of pressure inside both bubbles will decrease.

You will find many useful posts on different branches of Physics at your level at apphysicsresources.blogspot.com and at physicsplus.blogspot.com. The essential equations to be remembered in the section, ‘Electric field and Potential’ can be found here.