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

Two Kerala Engineering Entrance 2005 Questions on Electrostatics

I never did a day’s work in my life. It was all fun.

– Thomas Alva Edison

Here are two multiple choice questions which appeared in Kerala Engineering Entrance 2005 question paper:

(1) A soap bubble is charged to a potential of 16 V. Its radius is then doubled. The potential of the bubble now will be

(a) 16 V

(b) 8 V

(c) 4 V

(d) 2 V

(e) zero

The potential of a bubble of radius R carrying charge Q is Q/4πε₀R so that for a given charge the potential is inversely proportional to the radius. Therefore when the radius is doubled, the potential is halved. The answer is 8 V [Option (b)].

(2) A parallel plate capacitor of capacitance 10 μF is charged to 1μC. The charging battery is removed and then the separation between the plates is doubled. Work done during the process is

(a) 5 μJ

(b) 0.05 μJ

(c) 1 μJ

(d) 10 μJ

(e) 50 μJ

The work done is equal to the increase in the energy of the capacitor.

The initial energy is Q²/2C where C is the initial capacitance and Q is the charge.

Therefore initial energy, Q²/2C = 10^–12/(20×10^–6) = 5×10^–8 J = 0.05 μJ.

When the separation between the plates is doubled, the capacitance is halved (since the capacitance is Kε₀A/d with usual notations) and hence the energy is doubled. The final energy is thus 0.1 μJ.

The increase in energy, which is equal to the work done, is 0.05 μJ.

You will find all posts related to electrostatics on this site by clicking on the label ‘electrostatics’ below this post. More useful questions with solution on electrostatics can be found here as well as here.

Electrostatics: Two Questions (MCQ) on Sharing of Charge

(1) Three identical conducting spheres X, Y and Z carrying charges 8 μC, – 4.8 μC and 6.4 μC respectively are kept in contact and then separated from one another. Then, the sphere Y will have approximately

(a) an excess of 3×10¹³ electrons

(b) a deficiency of 3×10¹³ electrons

(c) a deficiency of 6×10¹³ electrons

(d) an excess of 6×10¹³ electrons

(e) a deficiency of 2×10¹³ electrons

The total charge on the spheres is (8 – 4.8 + 6.4) μC = 9.6 μC. Since the spheres are identical, they have the same capacitance and the charge is shared equally by them. Therefore, the charge on each sphere is 3.2 μC.

Since the charge on the sphere Y (and the spheres X and Z) is positive, there will be a deficiency of electrons.

Remembering that the electronic charge is 1.6×10^–19 coulomb, the deficiency of electrons on the sphere is 3.2×10^–6/1.6×10^–19 = 2×10¹³. Therefore the correct option is (e).

(2) Two capacitors C₁ and C₂ of capacitance 2 μF and 3 μF are charged to 50 volt and 40 volt respectively and arranged as shown, with the key K open. On closing the key, the charge flowing through the key will be

(a) 8 μC

(b) 12 μC

(c) 32 μC

(d) 64 μC

(e) 132 μC

You must remember that the total charge is conserved in all situations. In the above question, the total charge is C₁V₁ + C₂V₂ = 100 μC + 120 μC = 220 μC.

When the key K is closed, the capacitors get connected in parallel and the common potential difference between their terminals will be

V = (C₁V₁ + C₂V₂)/(C₁ + C₂) = 220 μC/ 5 μF = 44 volt.

The charge on C₁ after closing the key will be C₁V = 88 μC.

Since the initial charge on C₁ is 100 μC, the charge flowing through the key must be (100–88) μC = 12 μC.

[You will obtain the same result on considering C₂].

Questions on Electrostatics

The following question which appeared in IIT 1997 Entrance test paper checks whether you have a thorough understanding of the basic relation between the electric field and potential:
A non-conducting ring of radius 0.5m carries a total charge of 1.11×10^–10 C distributed non uniformly on its circumference producing an electric field E everywhere in space. The value of the line integral ∫ –E.dl between the limits l = ∞ to l = 0 ( l = 0 being the centre of the ring) in volts is
(a) +2 (b) –1 (c) –8 (d) zero
The line integral ∫ –E.dl between the limits l = ∞ and l = 0 gives the work done in bringing unit positive charge from infinity to the centre of the ring and therefore is equal to the electric potential at the centre of the ring. If the total charge on the ring is Q coulomb, the potential at the centre is V = (1/4πε₀)×Q/r where ‘r’ is the radius of the ring. Therefore, V = 9×10⁹ ×1.11×10^–10/0.5 = 2, very nearly. So, the correct option is (a).
Now, consider the following MCQ:
A dielectric slab (dielectric constant = K) of thickness ‘t’ is placed between the plates of a parallel plate air capacitor. If the capacitance of the capacitor is to be restored to the original value, the separation between the plates is to be increased by
(a) kt (b) k/t (c) t/k (d) t –(k/t) (e) t –(t/k)
When a dielectric slab of thickness ’t’ is introduced between the plates, the electric fields in the air space and in the dielectric space are respectively q/ε₀A and q/Kε₀A where ‘q’ is the charge on each plate (+q on one, –q on the other) and A is the area of the plate. The P.D. between the plates is V = (q/ε₀A)(d–t)+(q/Kε₀A)t = (q/ε₀A)[d–t+(t/K)]. The capacitance of the system on introducing the dielectric is C = q/V = ε₀A/[d–t+(t/K)] = ε₀A/[d–(t – t/K)].
Since the capacitance of the capacitor with air filling the entire space between the plates is ε₀A/d, the effect of introducing the dielectric slab of thickness ‘t’ is to reduce the thickness of air by t– (t/K). In order to restore the capacitance to the original value, the separation between the plates is to be increased by t– (t/K).
The following MCQ appeared in Kerala Engineering Entrance 2003 question paper:
A parallel plate capacitor has a capacitance of 100 pF when the plates of the capacitor are separated by a distance of ‘t’. Then a metallic foil of thickness t/3 is introduced between the plates. The capacitance will then become
(a) 100 pF (b) (3/2)100 pF (c) (2/3)100 pF (d) (1/3)100 pF (e) (1/2)100 pF
As shown in the previous discussion, the capacitance of a parallel plate capacitor with a dielectric slab of thickness ‘t’ between the plates separated by a distance ‘d’ is given by,C = ε₀A/[d–t+(t/K)]. In the present problem, ‘d’ is to be replaced by ‘t’ and ‘t’ is to be replaced by t/3. Further, the dielectric constant K is to be replaced by ∞ since the dielectric constant of a conductor is infinite. The capacitance therefore becomes ε₀A/(t–t/3) = (3/2) ε₀A/t = (3/2)100 pF (since the original capacitance with air alone as the dielectric is ε₀A/t = 100 pF).

You will find more multiple choice questions with solution at physicsplus: Questions on Electrostatics

Two Multiple Choice Questions from Electrostatics
Let us consider the following questions:
(1) What fraction of the energy drawn from the charging battery is stored ina capacitor?

(a) 100% (b) 50% (c)75% (d) 70.7% (e) 41.4%

With usual notations, the energy of a charged capacitor is ½CV^2 = ½QV. Since the charge Q is flowing under the action of the e.m.f. V of the charging battery, the energy supplied by the battery is QV. Therefore, the fraction of the energy stored is ½QV/QV = ½ = 50%.

(2) A charge ‘q’ is placed at the centre of the line joining two equal point charges, each equal to +Q. This system of three charges will be inequilibrium if q is equal to

(a) +Q (b) +Q/2 (c) –Q/2 (d) +Q/4 (e) –Q/4

+Q____q_____+Q Since the force on the charge ‘q’ placed at the centre will be zero irrespective of the sign of ‘q’, it is enough to set the condition for zero net force on the charge +Q. Therefore, 1/4πε0[Q^2/d^2 + Qq/(d/2)^2] = 0 where ‘d’ is the separation between the charges Q&Q. From this we get q = -Q/4, making (e) the correct option.