First slide

Capacitance

Question

A parallel plate capacitor of plate area A and plate separation d is charged to a potential V and then battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q E and If denote respectively, the magnitude of charge of each plate, the electric field between the plates (after the slab is inserted) and work done on the system in question in the process of inserting the slab, then

Moderate

A

$Q = \frac{ε_{0} AV}{d}$
B

$Q = \frac{ε_{0} KAV}{d}$
C

$E = \frac{V}{Kd}$
D

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

Solution

$C_{0} = \frac{ε_{0} A}{d} and Q_{0} = C_{0} V = \frac{ε_{0} AV}{d} . . . (1)$
When the battery is disconnected, the charge remains the same, i.e., $Q = ε_{0} AV / d$ . Electric field in the dielectric between plates of capacitor
$E = \frac{E_{air}}{K} = \frac{1}{K} (\frac{V}{d}) . . . . (2)$
After the insertion of dielectric, the new capacity C' becomes
   $C^{'} = K (\frac{ε_{0} A}{d}) = KC$
Now,    $V^{'} = \frac{Q}{C^{'}} = \frac{Q}{KC} = \frac{V}{K}$
Work done on the system
$\begin{array}{l} W = \frac{1}{2} {CV}^{2} - \frac{1}{2} C^{'} V^{' 2} \\ = \frac{1}{2} {CV}^{2} - \frac{1}{2} KC {(\frac{V}{K})}^{2} = \frac{1}{2} {CV}^{2} (1 - \frac{1}{K}) \\ = \frac{1}{2} (\frac{ε_{0} A}{d}) V^{2} (1 - \frac{1}{K}) \end{array}$

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