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 be energy stored in the capacitor be two times the initial energy stored?

• A. $l/4$
• B. $l/2$
• C. $l/3$
• D. $2l/3$

Answer 1

Answer: (C)

$C_i=\frac{{\epsilon }_{0}A}{d}$
$C_i=\frac{{\epsilon }_0\ell w}{d}$
$C_f=C_1+C_2$
$C_f=\frac{K{\epsilon }_0{A}_1}{d}+\frac{{\epsilon }_0{A}_2}{d}$
$C_f=\frac{K{\epsilon }_{0}wx}{d}+\frac{{\epsilon }_{0}w(\ell -x)}{d}$
$C_f=2C_i$
$\Rightarrow \frac{K{\epsilon }_{0}wx}{x}+\frac{{\epsilon }_{0}w(\ell -x)}{d}=\frac{2{\epsilon }_0\ell w}{d}$
$4x+\ell -x=2\ell$
$x=\frac{\ell }{3}$