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.

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.

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 Questions from Electrostatics

Here are the two questions from electrostatics which appeared in the All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 question paper:

(1) A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of the ring is:

(a) E along OK

(b) E along KO

(c) 3E along OK

(d) 3E along KO

Since the ring is conducting, the charge on it gets distributed uniformly along it. The resultant electric field at the centre of the ring is zero (since a unit positive test charge placed at the centre will be pushed by equal radial forces all around). The field due to the charge on the part ACDB of the ring is therefore equal and opposite to the field due to the charge on the part AKB of the ring. So the answer is E along OK [option (a)].

[It would have been better if it is mentioned that K is the mid point of the arc AKB].

(2) The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is

(a) ε₀ E²/Ad

(b) ε₀ E²Ad

(c) ½ ε₀E²Ad

(d) ½ ε₀E²/Ad

The energy (U) required for charging a capacitor of capacitance C to V volt is given by

U = ½ CV² where C = ε₀A/d for a parallel plate capacitor with air (or vacuum) as dielectric.

Since E = V/d, we have V = Ed

Therefore, U = (½) (ε₀A/d)(Ed)² = (½) ε₀E²Ad

You can find more questions (with solution) in this section at AP Physics Resources.

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.

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].

Two Questions (MCQ) on Electric Potential

(1) An infinite number of point charges each equal to +Q coulomb are arranged at random around a point P such that the distances of the charges from the point P are 1m, 2m, 4m, 8m, 16m,…….etc... The electric potential at P is

(a) zero (b) infinite (c) negligibly small

(d) Q/2πε₀ (e) Q/4πε₀

Note that the electric potential is a scalar quantity. Therefore, the direction of the location of the charge does not matter and the potentials simply add up. The resultant potential (V) at P is given by

V = (Q/4πε₀) × [(1/1) + (1/2) + (1/4) + (1/8) + ………]

The infinite series within the square bracket yields a value equal to 2 so that V = Q/2πε_0.

(2) A thin spherical conducting shell of radius R has a charge +Q. Another point charge –q is placed at the centre of the shell. The electrostatic potential at a point P distant R/2 from the centre of the shell is

(a) (Q/4πε₀R) – (2q/4πε₀R) (b) (q/4πε₀R) – (2Q/4πε₀R)

(c) – (2q/4πε₀R) (d) (Q/4πε₀R) (e) zero
The electrostatic potential at any point within the shell due to the charge Q on the shell is constant and is equal to Q/4πε₀R.

The potential at distance R/2 due to the charge –q placed at the centre of the shell is – q/4πε₀(R/2) = – 2q/4πε₀R.

Therefore, the net potential (V) at the point P distant R/2 from the centre of the shell is given by

V = (Q/4πε₀R) – (2q/4πε₀R), given in option (a).

[Note that positive charges will produce positive potential where as negative charges will produce negative potential].

You can find more posts on electrostatics by clicking on the label electrostatics below this post or on the left side of this page.

You will find similar multiple choice questions with solution at physicsplus also.

KEAM 2007(Medical) Question on Electric Dipole

The following MCQ which appeared in Kerala Medical Entrance 2007 question paper is straight forward, requiring your knowledge of the expression for torque on a dipole:

An electric dipole consists of two opposite charges each 0.05 μC separated by 30 mm. The dipole is placed in an uniform external electric field of 10⁶ NC^–1. The maximum torque exerted by the field on the dipole is

(a) 6×10^–3 Nm (b) 3×10^–3 Nm (c) 15×10^–3 Nm

(d) 1.5×10^–3 Nm (e) 9×10^–3 Nm

The torque on a dipole of moment p in an electric field E is the vector product p×E. The magnitude of the torque (τ) is pEsinθ where θ is the angle between dipole moment vector and the electric field vector. The maximum value of the torque is pE (when θ = 90º).

Note that p = q×2a where ‘q’ is the charge and 2a is the separation between the charges.

Therefore, maximum torque = (0.05×10^–6×30×10^–3) ×10⁶ = 1.5×10^–3 Nm.

Electrostatic Potential- An IIT-JEE 2007 MCQ

The following MCQ which appeared in IIT-JEE 2007 question paper is meant for checking whether you have the correct idea about electric potential and potential difference:

Positive and negative point charges of equal magnitude are kept at (0,0, α/2) and(0,0,–α/2), respectively. The work done by the electric field when another positive point charge is moved from (–α,0,0) to (0,α,0) is

(a) positive

(b) negative

(c) zero

(d) depends on the path connecting the initial and final positions

You should have a three dimensional mental view of the Cartesian coordinate system to work out this problem in no time. Even without drawing a figure, some of you will be able to work this out. But you may draw a quick figure to help you.

In the figure shown, the positions of the equal positive and negative charges and the movement of the third positive charge from P to Q are indicated. The path shown is straight, but it can be any curve, since the work done is independent of the path in an electrostatic field and is dependent on the initial and final positions only.

Since P and Q are equidistant from the positive and negative charges, the resultant potentials at P as well as Q due to these charges is zero. Therefore, no work need be done for moving the third charge from P to Q [Option (c)].

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