First slide

Capacitance

Question

A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then battery is disconnected. A slab of dielectric constant K is then inserted between the plates. If q, E and W denotes the magnitude of charge on each plate, electric field between the plates (After the slab insertion) and work done on the system, then

Moderate

A

$q = \frac{ε_{0} A V}{d}$
B

$E = \frac{V}{K d}$
C

$W = \frac{ε_{0} A V^{2}}{2 d} (\frac{1}{K} - 1)$
D

All of these

Solution

$C^{1} = K C = \frac{K ε_{0} A}{d}; V^{'} = \frac{q}{C^{'}} = \frac{V}{K}$
$q = become same = \frac{ε_{0} A V}{d}$
$E^{'} = \frac{V}{d} = \frac{V}{K d}$
$W = U_{f} - U_{i} = \frac{q^{2}}{2 C} [\frac{1}{K} - 1]$

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