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.