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

Kerala Medical Entrance (KEAM) 2009 Questions from Electrostatics

Keral Medical Entrance 2009 question paper contained three questions from electrostatics. They are simple except for the first one (given below) which will demand some three dimensional imagination.

(1) The total electric flux through a cube when a charge 8q is placed at one corner of the cube is

(a) ε₀ q

(b) ε₀ /q

(c) 4πε₀ q

(d) q/4πε₀

(e) q/ε₀

You should remember that the total electric flux originating from a charge q is q/ε₀.

[You will definitely remember that the electric field E at a point distant r from a point charge is q/4πε₀r². The electric field is equal to the electric flux through unit area held normal to the direction of the field. If you imagine a spherical surface with the charge q at the centre, the surface area of the sphere is 4πr² and the total flux passing normally through the spherical surface is (q/4πε₀r²) ×4πr² = q/ε₀]

In the question the charge 8q is placed at one corner of the cube. The total electric flux originating from this charge is 8q/ε₀. But only one-eigths of the flux passes through the cube. (You can place seven more cubes with their corners touching the charge to cover the entire volume around the charge).

The electric flux through a cube when a charge 8q is placed at one corner of the cube is therefore (1/8)×(8q/ε₀) = q/ε₀.

(2) A uniform electric field E exists along positive x-axis. The work done in moving a charge 0.5 C through a distance 2 m along a direction making an angle 60º with x-axis is 10 J. Then the magnitude of electric field is

(a) 5 Vm^–1

(b) 2 Vm^–1

(c) √5 Vm^–1

(d) 40 Vm^–1

(e) 20 Vm^–1

The force (F) acting on charge q in electric field E is given by F = qE along the direction of the field (along the positive x-axis here). Since the displacement (d) of the charge is inclined at 60º with the x-axis, the work done (W) is given by

W = qEd cos 60

Therefore, 10 = 0.5×E×2×½, from which E = 20 Vm^–1.

(3) A capacitor of capacitance C is charged to a potential V. If it carries charge Q, then the energy stored in it is

(a) ½ CV

(b) QV

(c) ½ QV²

(d) CV²

(e) ½ QV

The answer is ½ QV which you can obtain from ½ CV² which is the form most of you will usually remember:

½ CV² = ½ CV×V = ½ QV

[The capacitor is charged from zero potential to the potential V. The charge Q is therefore transferred to the capacitor at an average potential of (0+V)/2 = V/2. The work done is therefore QV/2.

To be more rigorous, if the charge on the capacitor at any instant of charging is q and the potential at the instant is v, the work done in transferring an additional charge dq is vdq= (q/C)dq. The total work done in transferring the entire charge Q is ₀ ∫^Q(q/C)dq = Q²/2C = ½ ×Q×(Q/C) = ½ QV].

You will find more questions (with solution) of various entrance examinations at physicsplus.blogspot.com.