Electrostatics - Multiple Choice Questions on Electric Field and Potential

Today we will discuss a few simple, but interesting, multiple choice questions on electrostatics. Here are the questions:

(1) ABC is an equilateral triangle. When positive charges Q and 2Q are placed at points A and B, the electric potential at the mid point (O) of AB is found to be 180 V. What is the electric potential at the vertex C of the triangle under this condition?

(a) 60 V

(b) 90 V

(c) 120 V

(d) 180 V

(e) 240 V

The distance of the vertex C of the triangle from the charges is twice the distance from the point O. Since the electric potential due to a point charge is inversely proportional to the distance, the potential at the vertex C must be half the value at C. So the answer is 90 V.

[Mathematically, you will write (with usual notations)

(1/4πε₀)(3Q/r) = 180 and

(1/4πε₀)(3Q/2r) = x

This gives x = 90 V]

(2) Equal and opposite point charges + Q and – Q are fixed at the corners A and B of an equilateral triangle (Fig.). If a negative point charge (– q) is placed at the vertex C of the triangle, the net force on this charge is directed

(a) towards right

(b) towards left

(c) towards A

(d) towards B

(e) towards the mid point of side AB

The force on the charge – q due to the charge + Q is attractive and is directed along CA. The force due to the charge – Q is repulsive and is directed along BC. Since these forces have the same magnitude, their resultant acts parallel to BA. The net force on the charge – q is thus directed leftwards [Option (b)].

(3) A cube of side 4 cm has a constant electric potential of 2 volt on its surface. If there are no charges inside the cube, the potential at a distance of 1 cm from the centre of the cube is

(a) zero

(b) 0.25 V

(c) 0.5 V

(d) 1 V

(e) 2 V

Since the surface of the cube is an equipotential surface and there are no charges inside the cube, the electric field inside the cube must be zero. This means that the electric potential everywhere inside the cube is 2 V itself [Option (e)].

(4) A spherical conductor of radius R has a central cavity (Fig.) with a negative point charge (–q) located at the centre of the cavity (without touching the surface of the cavity). The electric fiel at a point P distant r from the centre of the sphere is

(a) q/4πε₀r² directed towards the sphere

(b) q/4πε₀r² directed away from the sphere

(c) q/4πε₀r² directed upwards

(d) q/4πε₀r² directed downwards

(e) zero

The negative charge at the centre of the cavity will induce positive charge q on the surface of the cavity and negative charge –q on the outer surface of the sphere. The situation is similar to that of a spherical conductor carrying a charge –q so that the magnitude E of the electric field at the point P is given by

E = q/4πε₀r²

Since the field is due to negative charge, the direction of the field is towards the centre of the sphere.

Multiple Choice Questions on Unusual Combinations of Capacitors

"It is unwise to be too sure of one's own wisdom. It is healthy to be reminded that the strongest might weaken and the wisest might err."

– Mahatma Gandhi

Today we will consider a few questions on unconventional combinations of capacitors which occasionally find place in entrance examination question papers conducted for admitting students to medical, engineering and other degree courses.

(1) Four parallel metal plates are arranged in air as shown, with the same separation ‘d’ between neighbouring plates. If the area (of one surface) of each plates is A, the capacitance between the terminals T₁ and T₂ is

(a) 4ε₀A/d

(b) ε₀A/d

(c) 2ε₀A/d

(d) 3ε₀A/d

(e) ε₀A/3d

Let us number the plates starting from the top as 1, 2, 3, and 4 respectively. The lower surface of plate No.1 and the upper surface of plate No.2 with air in between makes a capacitor of capacitance ε₀A/d.

The lower surface of plate No.2 and the upper surface of plate No.3 with air in between makes another capacitor of capacitance ε₀A/d.

The lower surface of plate No.3 and the upper surface of plate No.4 with air in between makes a third capacitor of capacitance ε₀A/d.

This arrangement therefore makes three capacitors in parallel. The capacitance value of each capacitor is ε₀A/d so that the parallel combined value which appears between the terminals T₁ and T₂ is 3ε₀A/d [Option (d)].

(2) If the connection of metal plates in the above question is modified as shown in the adjoining figure, the capacitance between terminals T₁ and T₂ is

(a) ε₀A/2d

(b) 3ε₀A/2d

(c) 2ε₀A/d

(d) 4ε₀A/d

(e) 2ε₀A/3d

In this case the capacitor at the centre (having capacitance ε₀A/d) appears directly across the terminals T₁ and T₂. The upper and lower capacitors (each of value ε₀A/d) are connected in series and this series combination having effective capacitance ε₀A/2d also appears between the terminals T₁ and T₂. Therefore, the effective capacitance between T₁ and T₂ is ε₀A/d + ε₀A/2d = 3ε₀A/2d [Option (b)].

(3) If a fifth plate is added to the combination shown in question No.1, as shown in the figure, what is the effective capacitance between T₁ and T₂?

(a) ε₀A/4d

(b) ε₀A/5d

(c) 5ε₀A/d

(d) 4ε₀A/5d

(e) 4ε₀A/d

Once you answer question No.1, this question becomes quite simple. You can easily see that there are four identical capacitors (each having capacitance ε₀A/d) connected in parallel. Therefore, the effective capacitance between T₁ and T₂ is 4ε₀A/d [Option (e)].

You can find a few more questions (with solution) in this section here.

Multiple Choice Practice Questions on Electrostatics

“Live as if you were to die tomorrow. Learn as if you were to live for ever.”

– Mahatma Gandhi

If you grasp the fundamental principles well, you will be able to answer multiple choice questions within the stipulated time. If you are confused about the fundamental principles, you will be tempted to waste your time while dealing with even simple questions just because of some simple distractions. See the following questions:

(1) A battery of emf 12 V is connected in series with two initially uncharged capacitors, each of value 100 μF (Fig.). The capacitors are fully charged. If the total energy spent by the battery for charging the capacitors is E joule, the energy of each capacitor is

(a) E

(b) E/2

(c) 2 E

(d) E/4

(e) E/8

If the total charge supplied by the battery is Q coulomb, the total energy E supplied by the capacitor is given by

E = VQ joule where V is the emf of the battery (which is 12 volt).

The two identical 100 μF capacitors in series is equivalent to a single capacitor of capacitance C = 50 μF. The energy of this charged capacitor combination is ½ CV² joule = ½ VQ joule since Q = CV.

Therefore, the energy of the charged capacitor combination is ½ E.

Since there are two identical capacitors in the combination, the energy of each capacitor is E/4.

[So the capacitors have gained a total energy of E/2, even though the battery has delivered a total energy E. What happens to the other half of the energy? Well, it is irrecoverably lost as heat generated in the resistance of the circuit].

(2) Charges of equal magnitude are fixed at the diagonally opposite corners A and C of a non conducting square frame ABCD (Fig.). In case (i) the charges at A and C are positive and a third free negative charge q is moved from B to D. In case (ii) the charges at A and C are negative and a third free negative charge q is moved from B to D. In case (iii) the charges at A and C are of opposite sign and a third free negative charge q is moved from B to D. Pick out the correct statement regarding the work done for moving the charge q:

(a) In case (i) the work done is positive

(b) In case (ii) the work done is positive

(c) In case (i) the work done is negative

(d) In case (iii) the work done is positive

(e) In all cases the work done is zero

In case (i) the potentials at B and D are positive, but of equal value. In case (ii) the potentials at B and D are negative, but of equal value. In case (iii) the potentials at B and D are zero. Therefore, in all cases the potential difference between points B and D is zero so that the work done in moving the charge q from B to D is zero.

You will find some useful questions (MCQ) with solution here.

AIEEE 2009 Multiple Choice Questions (MCQ) with Solution on Electrostatics

The following four questions which appeared in the All India Engineering/Architecture Entrance Examination 2009 (AIEEE 2009) will be beneficial to most of the entrance test takers. Those who prepare for AP Physics C exam and physics GRE may take special note of question no.2. Here are the questions with solution:

(1) A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electric force on Q is zero, then Q/q equals

(1) – 1/√2

(2) – 2√2

(3) –1

(4) 1

The electric force F_QQ (Fig.) on Q because of the charge Q at the opposite corner is repulsive and is given by

F₁ = (1/4πε₀)(Q² /2a²) since the distance between the charges Q and Q is 2×a/√2 = a√2

The electric forces F_Qq and F_Qq (Fig.) on Q because of the charges q and q at the adjacent corners are perpendicular to each other. Each force has magnitude given by

F_Qq = (1/4πε₀)(Qq/a²).

Their resultant F₂ has magnitude √2 times the above value:

F₂ = (1/4πε₀)(√2 Qq/a²).

Since the net force on Q is zero, we have

F₁ + F₂ = 0

Therefore, (1/4πε₀)(Q² /2a²) + (1/4πε₀)(√2 Qq/a²) = 0

This gives Q/q = – 2√2

[The charges Q and q must be of opposite sign so that the force between Q and q is attractive to ensure that F₁ and F₂ are opposite in directions and the net force on Q is zero as given in the question].

(2) Let P(r) = Qr/πR⁴ be the charge density distribution for a solid sphere of radius R and total charge Q. For a point ‘p’ inside the sphere at distance r₁ from the centre of the sphere, the magnitude of electric field is

(1) Qr₁² /3πε₀ R⁴

(2) 0

(3) Q/4πε₀ r₁²

(4) Qr₁² /4πε₀ R⁴

Charge dQ contained in the spherical shell of radius r and thickness dr is (Qr/πR⁴)(4πr²dr) = (4Q/R⁴) r³dr

Charge Q₁ contained in the spherical volume of radius r₁ is therefore given by

Q₁ = ₀ ∫^r1 (4Q/R⁴) r³dr = (4Q/R⁴)(r₁⁴/4) = Qr₁⁴ /R⁴

If E is the electric field at distance r₁ from the centre of the sphere, we have from Gauss theorem,

4πr₁²E = Qr₁⁴ /R⁴ε₀ from which

E = Qr₁² /4πε₀ R⁴

[Remember that the charges distributed with spherical symmetry outside the point p will produce zero field at p and the field at p is indeed given correctly by the above expression].

(3) Two points P and Q are maintained at the potentials of 10 V and – 4 V, respectively. The work done in moving 100 electrons from P to Q is

(1) 2.24×10^–16 J

(2) – 9.60×10^–17 J

(3) 9.60×10^–17 J

(4) – 2.24×10^–16 J

This is a simple question. The potential difference ∆V between P and Q is 10 V – (– 4 V) = 14 V. Since the electrons are negatively charged, external work (positive work) has to be done to move them from the higher potential point P to the lower potential point Q.

The work done = q∆V = (100×1.6×10^–19)×14 J = 2.24×10^–16 J

(4) This question contains Statement-1 and statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement 1 : For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q.

Statement 2 : The net work done by a conservative force on an object moving along a closed loop is zero.

(1) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statment-1.

(2) Statment-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.

(3) Statement-1 is false, Statement-2 is true.

(4) Statement-1 is true, Statement-2 is false

Electrostatic field is conservative and the net work done by an electrostatic field on a charged particle is dependent only on the initial and final positions of the charged particle. (It is independent of the path connecting the initial and final positions).

The correct option is (1).

You will find some more useful multiple choice questions with solution from electrostatics at AP Physics Resources.

IIT-JEE 2008 Multiple Choice Questions on Electrostatics

The following two questions from Electrostatics were included under Straight Objective Type (Multiple choice, single answer type) in the IIT-JEE 2008 question paper:

(1) Consider a system of three charges q/3, q/3 and –2q/3 placed at points A, B and C respectively as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB = 60º

(a) The electric field at point O is q/8πε₀R²

(b) The potential energy of the system is zero

(c) The magnitude of the force between the charges at C and B is q²/54πε₀R²

(d) The potential at point O is q/12πε₀R

The electric field at O is due to the negative charge at C only since the equal positive charges situated at A and B will produce equal and opposite fields at O (and they will mutually cancel). The field at O is therefore negative and the option (a) is obviously wrong. Option (b) also is obviously wrong.

The magnitude of the force (F) between the charges at C and B is given by

F = (1/4πε₀) [(2q/3)(q/3)/(2Rsin60º)²].

Thus F = q²/54πε₀R² as given in option (c).

(2) A parallel plate capacitor with plates of unit area and separation d is filled with a liquid of dielectric constant K = 2. The level of liquid is d/3 initially. Suppose the liquid level decreases at a constant speed V. The time constant as a function of time t is

(a) 6ε₀R/(5d + 3Vt)

(b) (15d + 9Vt) ε₀R/(2d² – 3dVt – 9V²t²)

(c) 6ε₀R/(5d – 3Vt)

(d) (15d – 9Vt) ε₀R/(2d² + 3dVt – 9V²t²)

When a dielectric slab of thickness t₁ and dielectric constant K is introduced in between the plates of a parallel plate air capacitor of plate area A and plate separation d, the effective capacitance changes from ε₀A/d to ε₀A/[d–t₁ + (t₁/K)]

The effective capacitance C of a parallel plate capacitor with parallel dielectric slabs of thickness t₁, t₂, t₃ etc. of dielectric constants K₁, K₂, K₃ etc. respectively is given by the series combined value of capacitors with these dielectrics. Therefore,

1/C = 1/(K₁ε₀A/t₁) + 1/(K₂ε₀A/t₂) + 1/(K₃ε₀A/t₃) + etc.

With a single slab of thickness t₁ and dielectric constant K introduced between the plates, we have

1/C = 1/(Kε₀A/t₁) + 1/[ε₀A/(d – t₁)] from which C = ε₀A/[d–t₁ + (t₁/K)]

In the present problem, the thickness of the layer of liquid (which serves as the dielectric slab) at the instant t is (d/3) – V t and the thickness of the air space is (2d/3) + V t.

The effective capacitance is therefore given by

C = ε₀A/[{(2d/3) + Vt} + {(d/3) – Vt}/k)]

Substituting for A = 1 and K = 2, we obtain C = 6ε₀/(5d + 3Vt).

The time constant of the circuit = RC = 6ε₀R/(5d + 3Vt).

Reality is merely an illusion, albeit a very persistent one. 

–Albert Einstein 

Two Questions from Electrostatics-- Capacitor Combinations

Here are two multiple choice questions on effective capacitance of capacitor networks. [The first question was found to be some what difficult for the students (from their response)]:

(1) An unusual type of capacitor is made using four identical plates P₁, P₂, P₃ and P_4, each of area A arranged in air with the same separation d as shown in the figure. A thin wire outside the system of plates connects the plates P₂ and P₄. Wires soldered to the plates P₁ and P₃ serve as terminals T₁ and T₂ of the system. What is the effective capacitance between the terminals T₁ and T₂?

(a) (2ε₀A)/3d

(b) (3ε₀A)/2d

(c) (3ε₀A)/d

(d) (ε₀A)/d

(e) (ε₀A)/2d

The arrangement contains 3 identlcal capacitors C₁, C₂ and C₃ each of capacitance (ε₀A)/d arranged as shown in the adjoining figure. Plate P₂ is common for C₁ and C₂. Likewise, plates P₃ is common for C₂ and C₃. The capacitors C₂ and C₃ are connected in parallel (to produce an effective value 2ε₀A/d) and this parallel combination is connected in series with the capacitor C₁ of value ε₀A/d . The effective capacitance C between the terminals T₁ and T₂ is therfore given by

C = [(2ε₀A/d) ×(ε₀A/d)] / [(2ε₀A/d) +(ε₀A/d)].

Therefore, C = (2ε₀A)/3d

(2) Four 2 μF capacitors and a 3 μF capacitor are connected as shown in the figure. The effective capacitance between the points A and B is
(a) 3 μF

(b) 2 μF

(c) 1.5 μF

(d) 1 μF

(e) 0.5 μF

This is a very simple question once you identify the circuit to be a balanced Wheatstone’s bridge. The 3 μF capacitor is connected between equipotential points and you can ignore it. Without the diagonal branch, there are four 2 μF capacitors only. The capacitors in the upper pair are in series and produce a value of 1 μF. The capacitors in the lower pair also produce a value of 1 μF. Since they are in parallel the effective capacitance between the points A and B is 2 μF.