Question

Two parallel plate capacitors $A$ & $B$ have the same separation $d=8.85\times {10}^{-4}m$ between the plates. The plate areas of $A$ & $B$ are $0.04m^2$ & $0.02m^2$ respectively. A slab of di-electric constant (relative permitivity) $K=9$ has dimensions such that it can easily fill the space between the plates of capacitor $B$.
(i) The di-electric slab is placed inside $A$ as shown in the figure $\left(a\right)A$ is then charged to a potential difference of $110volt$. Calculate the capacitance of $A$ and the energy stored in it.
(ii) The battery is disconnected & then the di-electric slab is removed from $A$. Find the work done by the external agency in removing the slab from $A$.
(iii) The same di-electric slab is now placed inside $B$, filling it completely. The two capacitors $A$ & $B$ are then connected as shown in Figure $\left(c\right)$. Calculate the energy stored in the system.

Answer 1

$\left(i\right)$ Capacitor $A$

Plate area $A$$=0.04m^2$

Plate area $B$$=0.02m^2$

As per the given case,

Capacitor $A$ is combination of two capacitors $C_K$ and $C_o$ in parallel.

$\Rightarrow C_A=C_K+C_o$

$=\frac{K{\epsilon }_o{A}_1}{d}+\frac{K{\epsilon }_o{A}_2}{d}$

$=\frac{885\times {10}^{-12}}{8.85\times {10}^{-4}}[10\times 0.02]$

$\Rightarrow C_A=2\times {10}^{-9}F$

Energy stored in capacitor $=U_A=\frac{1}{2}{C}_{A}V2$

$\frac{1}{2}\times 2\times {10}^{-9}\times {\left(110\right)}^2$

$=1.21\times {10}^{-5}J$

$\left(ii\right)$ Work done in removing slab

$W=U_A^{\prime }-U_A$

Where,

$U_A^{\prime }$ Energy of capacitor without slab

Charge $Q=C_{A}V=2\times {10}^{-9}\times 110=2.2\times {10}^{-7}C$

Capacitor $C_A^{\prime }=\frac{{\epsilon }_o(A_1+A_2)}{d}$

$=\frac{8.85\times {10}^{-12}(0.02+0.02)}{8.85\times {10}^{-4}}$

$=0.4\times {10}^{-9}F$

$W=6.05\times {10}^{-5}-1.21\times {10}^{-5}=4.84\times {10}^{-5}J$

$\left(iii\right)$ Energy store when $A$ and $B$ are connected $=U$

$C_B=\frac{K{\epsilon }_o{A}_1}{d}=\frac{9\times 8.85\times {10}^{-12}\times 0.02}{8.85\times {10}^{-4}}=1.8\times {10}^{-9}F$

Capacitor $A$ and $B$ are in parallel.

$C=C_A^{\prime }+C_B^{\prime }$

$C=(0.4\times {10}^{-9})+(1.8\times {10}^{-9})=2.2\times {10}^{-9}F$

Energy tored in system $=U=\frac{Q^2}{2C}$

$=\frac{{(2.2\times {10}^{-7})}^2}{2\times 2.2\times {10}^{-9}}$

$=1.1\times {10}^{-5}J$