First slide

Capacitance

Question

The capacity of a parallel plate condenser is 10 $μ$ F when the distance between its plate is 8 cm. If the distance between the plates is reduced to 4 cm, the capacity will be

Easy

A

$10 μF$
B

$5 μF$
C

$20 μF$
D

$40 μF$

Solution

$C = \frac{ε_{0} A}{d} or 10 = \frac{ε_{0} A}{8}$
When the distance becomes d' , than new capacity
$C^{'} = \frac{ε_{0} A}{d^{'}} or C^{'} = \frac{ε_{0} A}{4} ∴ \frac{C^{'}}{10} = \frac{8}{4} or C^{'} = \frac{80}{4} = 20 μF$

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