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?

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$l / 4$

$l / 2$

$2 l / 3$

$l / 3$

answer is D.

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Detailed Solution

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