First slide

Capacitance

Question

The distance between the plates of a parallel plate Condenser is 4mm and potential difference is 60olts. If the distance between the plates is increased to12mm, then

Easy

A

The potential difference of the condenser will become 180 volts
B

The PD. will become 200 volts
C

The PD. will remain unchanged
D

The charge on condenser will reduce to one third

Solution

For an isolated capacitor charge and hence electric field remains unchanged on changing the separation so $E = V / d = Constant. Or \frac{V_{1}}{V_{2}} = \frac{d_{1}}{d_{2}}$
$\Rightarrow V_{2} = \frac{V_{1} \times d_{2}}{d_{1}} = \frac{60 \times 12}{4} = 180 V$

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