First slide

Insertion of dielectric in capacitor

Question

A parallel plate capacitor has plate of length ‘ l’ , width ‘w’ and separation of plates is ‘d’. It is connected to a battery of emf V. A dielectric slab of the same thickness ‘d’ and of dielectric constant k = 4 is being inserted between the plates of the capacitor. At what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored?

Moderate

A

$l / 4$
B

$l / 2$
C

$2 l / 3$
D

$l / 3$

Solution

Energy stored in capacitor (i) $V_{1} = \frac{1}{2} C_{1} V^{2} = \frac{1}{2} \frac{ε_{0} A}{d} V^{2} = \frac{1}{2} \frac{ε_{0} l w}{d} V^{2}$
$\begin{array}{l} S i m i l a r l y i n (i i), \\ p a r t s w i t h a i r a n d d i e l e c t r i c c a n b e c o n s i d e r e d i n p a r a l l e l \\ U_{2} = \frac{1}{2} C_{2} V^{2} = \frac{1}{2} (\frac{ε_{0} A_{1}}{d} + \frac{ε_{0} A_{2}}{d}) V^{2} = \frac{1}{2} (\frac{ε_{0} (l - x) w}{d} + \frac{4 ε_{0} x w}{d}) V^{2} \end{array}$
$\begin{array}{l} & a l s o g i v e n \\ U_{2} = 2 U_{1} \\ \Rightarrow (l - x) + 4 x = 2 l \\ \Rightarrow x = l / 3 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App