- Home
- Standard 12
- Physics
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$?$
$\frac{l}{4}$
$\frac{l}{2}$
$\frac{l}{3}$
$\frac{2l}{3}$
Solution
Before inserting slab
$C_{i}=\frac{\varepsilon_{0} A}{d}$
$C_{i}=\frac{\varepsilon_{0}(w)}{d}$
After inserting dielectric slab
$C_{f}=C_{1}+C_{2}$
$C_{f}=\frac{K \varepsilon_{0} A_{1}}{d}+\frac{\varepsilon_{0} A_{2}}{d}$
$C_{f}=\frac{K \varepsilon_{0} W x}{d}+\frac{\varepsilon_{0} w(\ell-x)}{d}$
$C_{f}=2 C_{i} \Rightarrow \frac{K{\varepsilon_0} w x}{x}+\frac{\varepsilon_{0} w(\ell-x)}{d}=\frac{2 \varepsilon_{0}(w)}{d}$
$4 x+\ell-x=2 \ell$
$x=\frac{\ell}{3}$
