A Parallel Plate Capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles then,
a) the charge on the capacitor increases
b) the voltage across the plate increases
c) the capacitance increases
d) the electrostatic energy stored in the capacitor increases

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2015-04-01T22:53:31+05:30

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Since battery is disconnected and distance is increased by insulating handles, the charge remains unchanged. So (a) is wrong.

Capacitance(C) is inversely proportional to the distance between the plates, d. As d is increased, C will decrease. So option (c) is wrong.

q=CV. Since q is constant and C is decreasing, V will increase. So option (b) is correct.

Change in Energy = 0.5qΔV. Since ΔV is positive, Energy of the system increases. So option (d) is correct.

So b and d are correct.
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2015-04-01T23:45:15+05:30

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Q = C * V
     Q  = charge on the capacitor plates (each)
     V = voltage across the plates
     C = capacitance of the capacitor

C = ε A / d  ,  Where A = area of each of the plates
                         d = distance between the plates.
                         ε = permittivity of the medium in between the plates.

 If the distance d increases, then the capacitance C decreases.  As a consequence, the charge on the capacitance is not affected.  So Voltage difference between the plates increases.

Electrostatic energy = 1/2 C V² = 1/2 Q V = Q² / 2 C
     So as the capacitance decreases, the electrostatic energy increases.  The work done in moving the capacitor plates against electrostatic attraction, is stored in the capacitor as electrostatic potential energy.

So,  b  and  d  are true.

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