Question

A parallel plate capacitor has a dielectric slab (length l, width b and thickness d) in it. The capacitor is charged by a battery and then the battery is disconnected. Now, the slab is started to pull out slowly at t=0, with velocity v. At any time t, capacitance of the capacitor is :

• A. $\frac{{\in }_{0}b}{d}[kl-(k-1\left)vt\right]$
• B. $\frac{{\in }_{0}l}{d}[kl-kvt]$
• C. $\frac{{\in }_{0}b}{d}\left[\right(k-1\left)vt\right]$
• D. $\frac{{\in }_{0}b}{b}[l-(k-1\left)vt\right]$

Answer 1

The correct option is A $\frac{{\in }_{0}b}{d}[kl-(k-1\left)vt\right]$

Let the slab pull out by a distance $x$ at time $t$. So $x=vt$
Now there are two capacitors and they are in parallel.
$C_1=\frac{A_1{ϵ}_0}{d}=\frac{\left(xb\right){ϵ}_0}{d}$ and $C_2=\frac{A_{2}k{ϵ}_0}{d}=\frac{\left[\right(l-x\left)b\right]k{ϵ}_0}{d}$
Total capacitance , $C=C_1+C_2=\frac{\left(xb\right){ϵ}_0}{d}+\frac{\left[\right(l-x\left)b\right]k{ϵ}_0}{d}=\frac{b{ϵ}_0}{d}[x+kl-kx]=\frac{b{ϵ}_0}{d}[kl-(k-1\left)vt\right]$ as $x=vt$