Multiple Choice Questions on Electrostatics

“Example isn't another way to teach, it is the only way to teach.”

– Albert Einstein

Today we will discuss a few questions from electrostatics. You will find many questions (with solution) in this section discussed earlier on this site. You can access all those questions by clicking on the label ‘electrostatics’ below this post.

(1) A 6 μF capacitor is connected in series with a 2 μF capacitor. The 6 μF capacitor can withstand a maximum voltage of 3 kV where as the 2 μF capacitor can withstand a maximum voltage of 6 kV. The maximum voltage that the parallel combination can withstand is

(a) 2 kV

(b) 3 kV

(c) 6 kV

(d) 8 kV

(e) 12 kV

We have capacitors C₁ and C₂ (let us say) having values 6 μF and 2 μF. If the maximum voltage that the parallel combination can withstand is V_max, the voltage V₁ across the 6 μF capacitor on applying this voltage across the series combination is given by

            V₁ = V_max C₂/(C₁ + C₂)

[The charge Q on each capacitor on connecting the voltage V_max across the series combination is given by Q = C₁C₂ V_max/(C₁ + C₂), remembering that the effective capacitance of the series combination is C₁C₂/(C₁ + C₂)].

Therefore, V₁ = V_max×2/(6+2) = V_max/4

Since C₁ can withstand a maximum voltage of 3 kV we have V_max/4 = 3 kV

This gives V_max = 12 kV.

[Do not jump to a conclusion at this stage, You have to check whether C₂ will be intact on applying the above 12 kV across the series combination].

The voltage V₂ across the 2 μF capacitor on applying this voltage V_max across the series combination is given by

            V₂ = V_max C₁/(C₁ + C₂)

Threfore V₂ = V_max×6/(6+2) = 6 V_max/8

Since C₂ can withstand a maximum voltage of 6 kV we have 6 V_max/8 = 6 kV

This gives V_max = 8 kV.

This value of V_max being lower than that obtained (12 kV) on considering the 6 μF capacitor, the correct option is 8 kV.

(2) A uniform electric field of intensity E newton/coulomb directed along the positive x-direction exists in a region of space (Fig.). The x- direction is horizontal. A, B, C and D are points at the corners of a square of side a, with AB and CD  parallel to the x-direction. If the electric potential at point A is V volt, what is the potential (in volt) at the point D?

(a) V

(b) V – aE

(c) V – √2 aE

(d) V + √2 aE

(e) V + aE

Since the electric field acts along the positive x-direction, the potential decreases as we move along the positive x-direction.

[Remember that the electric field is the negative gradient of potential].

While moving from A to D the x-coordinate increases by ‘a’ and hence the potential decreases by aE. Therefore, the potential at D is V – aE [Option (b)].

(3) In the above question what is the potential difference between points A and C?

(a) V

(b) V – aE

(c) V + aE

(d) aE

(e) Zero

Since the electric field acts along the x-direction, the potential will change only if the x-coordinate changes. Points A and C have the same x-coordinates and hence they are at the same potential. Therefore, the potential difference between points A and C is zero.

(4) Two small identical spheres are charged equally and suspended in air by strings of equal length. The strings make a small angle θ with each other (Fig.). When the spheres are immersed in oil of density 800 kg m^–3 the angle between the strings is found to be unaltered. If the density of the material of the spheres is 1200 kg m^–3, what is the dielectric constant of the oil?

(a) 1.5

(b) 2.5

(c) 3

(d) 3.5

(e) 4

The repulsive electrostatic force F  between the spheres in air is given by

             F = (1/4πε₀) (q²/d²) where ε₀ is the permittivity of free space (and air, very nearly), qis the charge on each sphere and d is the distance between the spheres.

            When the spheres are in the oil the electrostatic force F₁ between the spheres is given by

            F₁ = (1/4πε₀K) (q²/d²) where K is the dielectric constant of the oil.

The real weight W of each sphere is given by

            W = Vρg where V is the volume, ρ is the density of the material of the sphere and  g is the acceleration due to gravity.

The apparent weight W₁ of each sphere when immersed in oil is given by

            W₁ = Vρg – Vσg where σ is the density of the oil.

[Note that Vσg is the upthrust or the force of buoyancy due to the oil]

When the spheres are in air, we have (Fig.)

     tan α = F /W = (1/4πε₀) (q²/d²)/ Vρg………………..(i)

When the spheres are in oil, we have

tan α = F₁ /W₁ = (1/4πε₀K) (q²/d²) / (Vρg – Vσg)……(ii)

Dividing Eq. (i) by Eq. (ii) we have

            1 = K(ρ – σ) / ρ

Therefore, K = ρ/(ρ – σ) = 1200/400 = 3

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.

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.

MCQ from Electrostatics- Charged Spherical Shells

Two thin concentric spherical shells of copper having radii R₁ and R₂ (R₂ > R₁) carry charges Q₁ and Q₂ respectively. The potential at a point P (Fig) distant ‘r’ from the common centre and located in between the shells is

(a) (1/4πε₀)(Q₁/r)

(b) (1/4πε₀)(Q₁ + Q₂)/r

(c) (1/4πε₀)[(Q₁/R₁)+ (Q₂/R₂)]

(d) (1/4πε₀)[(Q₁/R₁) + (Q₂/r)]

(e) (1/4πε₀)[(Q₁/r) + (Q₂/R₂)]

Potentials due to the two charged shells get added to give the net potential at P. The potential at P due to the smaller shell of radius R₁ is (1/4πε₀)(Q₁/r) since the point P is outside the shell at distance ‘r’ from its centre. The potential at P due to the larger shell of radius R₂ is (1/4πε₀)(Q₂/R₂) since the point P is inside this shell and the potential everywhere inside is the same as the potential at the surface of the shell. The net potential at P is obtained by adding these two potentials.

Therefore, the net potential at P = (1/4πε₀)[(Q₁/r) + (Q₂/R₂)]

[This question can be modified by making one of the charges or both charges negative. You will then have to substitute the charges with the proper sign in the above equation].

What is the electric potential at a point inside the smaller shell?

The point is then inside both shells and both produce constant potentials (1/4πε₀)(Q₁/R₁) and (1/4πε₀)(Q₂/R₂) respectively. The net potential is therefore (1/4πε₀)[(Q₁/R₁)+ (Q₂/R₂)]

What is the electric potential at a point outside the larger shell?

The point is then outside both shells and they produce potentials (1/4πε₀)[(Q₁/r) and (1/4πε₀)[(Q₂/r) respectively so that the net potential is (1/4πε₀)(Q₁ + Q₂)/r. This is a very simple case since you can treat the entire charge Q₁ + Q₂ to be located at the common centre of the shells.

Now, one more question: What is the electric field at a point P in between the two shells?

The larger shell does not produce any field at P since it is inside and there is no potential gradient inside. The field at P is therefore due to the smaller shell only and is equal to (1/4πε₀)[(Q₁/r²)

You can find more posts on electrostatics by clicking on the label 'electrostatics' below this post or by performing a search for 'electrostatics' making use of the blog search box on this page.

Charged Particles in Magnetic and Electric Fields

Here is a multiple choice question which appeared in AIEEE 2002 question paper:
If an electron and a proton having same momenta enter perpendicular to a magnetic field, then
(a) curved path of electron and proton will be same (ignoring the sense of revolution)
(b) they will move undeflected
(c) curved path of electron is more curved than that of the proton
(d) path of proton is more curved
The radius of the circular path of the electron is obtained by equating the magnetic force to the centripetal force: qvB = mv²/r. The radius ‘r’ is therefore given by r = mv/qB. The radius is therefore directly proportional to the momentum (mv) and inversely proportional to the charge (q) of the particle.

The momenta are given as equal in the problem. Since the proton and the electron have the same charge magnitudes and are moving in the same magnetic field B, they will follow paths of the same radius[Option (a)].
You might have noted that the path of a charged particle in an electric field is generally parabolic. This is because of the fact that in a uniform electric field, the electric force on the particle has the same direction everywhere. The motion is similar to the projectile motion in a gravitational field. Now, consider the following question:
A particle of charge ‘+q’ and mass ‘m’ is projected with a velocity ‘v’at an angle ‘θ’ with respect to the horizontal, in an electric field ‘E’ which is directed vertically downwards. If there are no gravitational or magnetic fields, the horizontal range of the particle is
(a) (v²sin2θ)/E (b) (v²sin2θ)/qE (c) (mv²sin2θ)/E (d) (mv²sin2θ)/qE (e) (qmv²sin2θ)/E

In the case of the motion of a projectile in a gravitational field, the expression for horizontal range is R = (v² sin2θ)/g. In the present case, the gravitaional acceleration ‘g’ is replaced by the acceleration produced by the electric field.
Acceleration produced by the electric field = Force/ Mass = qE/m. The correct option therefore is (d).
The following MCQ appeared in AIIMS 2004 question paper:
The cyclotron frequency of an electron gyrating in a magnetic field of 1 T is approximately
(a) 28 MHz (b) 280 MHz (c) 2.8 GHz (d) 28 GHz
The cyclotron frequency is the frequency with which a charged particle describes circular path in a magnetic field and is given by f =qB/2πm with usual notations. [You can get it this way: qvB = mrω² where ω is the angular frequency. Substituting v = ωr in this, we get ω = qB/m. Frquency f = ω/2π =qB/2πm].

Substituting for the mass and charge of the electron, we have

f = (1.6×10^–19 ×1)/ (2π×9.1×10^–31 ) = 28×10⁹ Hz = 28 GHz.

Multiple Choice Questions on Electrostatics:

The following M.C.Q. appearered in the question paper of A.I.E.E.E.2004 (It may appear to be time consuming on the first reading, but it is simple):Two spherical conductors B and C having equal radii and carrying equal charges on them repel each other with a force F when kept apart at some distance. A third spherical conductor A having the same radius as that of B but uncharged is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is
(a) F/4 (b) 3F/4 (c) F/8 (d) 3F/8If Q is the initial charge on B and C and ‘r’ is the distance between their centres, the initial repulsive force between B and C is F= (1/4πε₀)(Q²/ r²). When the uncharged sphere A is brought in contact with B, they share the charge Q and each will have a charge Q/2 since they are identical spheres. When the sphere A is then brought in contact with C they will share the total charge Q + Q/2 = 3Q/2 equally and each will have a charge 3Q/4. On removing the sphere A we have sphere B with charge Q/2 and the sphere C with charge 3Q/4. The new repulsive force between B and C is evidently (1/4πε₀)(Q/2)(3Q/4)/r² = (1/4πε₀)3Q²/8r² = 3F/8 [Option (d)].
Take a special note of the following M.C.Q. which often finds a place in entrance test papers: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 in equilibrium if 'q' is equal to
(a) +Q (b) +Q/2 (c) –Q/2 (d) +Q/4 (e) –Q/4
+Q____ q_____+Q

The force on the charge ‘q’ placed at the centre will be zero irrespective of the sign of ‘q’. For the system to be in equilibrium, the net force on the charge +Q should be zero. Therefore, (1/4πε₀)[Q² /d² + Qq/(d/2)²] = 0 where ‘d’ is the separation between the charges Q&Q. From this we get q = -Q/4 [Option (e)].The following M.C.Q is quite simple because of the symmetry in the arrangement of charges. Similar questions can be seen in entrance test papers:Equal charges of value -Q each are arranged at the eight vertices of a non-conducting skeleton cube of side ‘a’. If a point charge +Q is placed at the centre of the cube, the electrostatic force exerted by the eight negative charges on the positive charge at the centre is
(a) Q²/3πε₀a² (b) 4Q²/3πε₀a² (c) 8Q²/3πε₀a² (d) 16Q²/3πε₀a² (e) zero
The net electrostatic force on the charge Q at the centre is zero since the force on Q due to each negative charge is balanced by the force due to the negative charge at the diagonally opposite corner of the cube.
The following M.C.Q. which appeared in the A.I.E.E.E.2004 question paper is worth noting:
Four charges equal to –Q are placed at the four corners of a square and a charge ‘q’ is at the centre.If the system is in equilibrium, the value of ‘q’ is
(a) –Q(1+2√2)/4 (b) Q(1+2√2)/4 (c) –Q(1+2√2)/2 (d) Q(1+2√2)/2The system will be in equilibrium if the attractive force on the charge –Q at a corner due to the central charge ‘q’ is balanced by the net repulsive force on the same charge –Q due to the remaining three charges (-Q each) at the remaining three corners. Hence we have (if ‘a’ is the side of the square),
(1/4πε₀) [Qq/(√2 a/2)²] = (1/4πε₀) [Q²/(√2 a)²] + √2 × [(1/4πε₀) ×(Q²/a²)].The term on LHS is the attractive force between –Q and q which are separated by the distance half of √2 a. The first term on RHS is the repulsive force between –Q and the diagonally opposite charge –Q (which are separated by √2 a). The second term on RHS is the net repulsive force between –Q and the remaining two charges (-Q each).
Hence we obtain 4q = Q + Q(1+2√2) from which q = Q(1+2√2)/4 [Option (b)].

The following question was asked at the Kerala Medical Entrance test of 2006:Three identical charges each of 2μC are placed at the vertices of a triangle ABC. If AB+AC=12cm and AB.AC=32cm², the potential energy of the charge at A is
(a) 1.53J (b) 5.31J (c) 3.15J (d) 1.35J (e) 3.51J
The electrostatic potential energy of the charge at A ia (1/4πε0)(Q.Q/AB + Q.Q/AC) since the charges are of the same value, Q=2μC. Since AB.AC=32, AB = 32/AC. Substituting this in the equation, AB + AC = 12, we obtain 32/AC + AC = 12. Rearranging, (AC)² – 12AC + 32 = 0, which yields AC = 8 or 4. Since AB.AC = 32 cm², if AC = 8 cm, AB = 4 cm and vice versa.
Therefore potential energy = (1/4πε0)[(4×10^-12)/(4×10^-2) +(4×10^-12)/(8×10^-2)] joule.
Since 1/4πε0 =9×10⁹, the potential energy works out to 1.35 J, given in option (d).
Now consider the following M.C.Q.:Two isolated copper spheres of radii 3cm and 6cm carry equal positive charges of 30 units each. If they are connected by a thin copper wire and then the wire is removed, what will be the charge on the smaller sphere?
(a) 60 units (b) 40 units (c) 30 units (d) 20 units (e) 10 unitsThe total charge (Q1+Q2) in the system is 60 units. Since the connecting wire is thin, its capacitance can be neglected. Even though the potentials of the spheres are different initially, their potentials will become the same when they are connected by the copper wire.
The capacitance of a spherical conductor being 4πε0 r, the charges on the spheres will be directly proportional to their radii, since Q = CV and V is the same. Therefore we have Q1/Q2 = 3/6 = 1/2 and Q1+Q2 = 60. So, Q1 = 60×1/3 = 20 units. The potential of the spheres will change after removing the connecting wire; but the charges on them will not change. So, the correct option is (d).Let us discuss another M.C.Q.:
An infinite number of charges each equal to –q coulomb are placed on a straight line at x = 1m, 2m, 4m, 8m,16m,…….. What will be the potential at x = 0 due to these charges?
(a) infinite (b) 2q/4πε0 (c) –q/4πε0 (d) –2q/4πε0 (e) zeroThe potential is certainly negative since a negative charge will produce negative potential only. You should supply the sign of the charge in the expression for the net potential. The potential at x = 0 is given by, V = (-q/4πε0)(1+ ½ + ¼ + 1/8 + 1/16 + ………). Since the infinite series yields a value equal to 2, the answer is –2q/4πε0.

Very simple questions are often likely to mislead even very bright students. Be careful. Here is a very simple question:Nine negative charges, each of magnitude Q are arranged symmetrically along the circumference of a circle of radius R. The electric field at the centre of the circle is
(a) (1/4πε0) Q/R² (b) (1/4πε0) (9Q/R²) (c) -(1/4πε0) (9Q/R²)

(d) -(1/4πε0) (9Q/R²) cos(2π/9) (e) zero

When you place a positive test charge at the centre of the circle, it will experience zero net force since it is pulled equally by the nine charges arranged symmetrically all around. Therefore, the electric field at the centre is zero.

You will find all the posts on Electrostatics in this site by clicking on the label ‘electrostatics’ below this post.