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