Question

Three identical large metal plates of area $A$ are separated by distances $d$ and $2d$ from each other, as shown in the figure. The top metal plate is uncharged, while other metal plates have charges $+Q$ and $–Q$. Top and bottom metal plates are connected by switch $S$ through a resistor of unknown resistance. What is the energy (in $\text{mJ}$) dissipated in the resistor when the switch is closed and steady state is achieved?

Given: $\frac{{ϵ}_{0}A}{d}=6\mu \text{F},Q=60\mu \text{C}$

Answer 1

Given, $\frac{{ϵ}_{0}A}{d}=6\mu \text{F}$ and $Q=60\mu \text{C}$

The above circuit, when switch $S$ is closed, can be redrawn as -

The capacitance of the given system of plates are,

$C_1=\frac{{ϵ}_{0}A}{2d}=3\mu \text{F}$

$C_2=\frac{{ϵ}_{0}A}{d}=6\mu \text{F}$

When switch $S$ is closed, their potential becomes equal when steady state is achieved,

$\frac{q_1}{C_1}=\frac{q_2}{C_2}$

$\Rightarrow \frac{q_1}{3}=\frac{q_2}{6}$

$\Rightarrow q_2=2q_1$

Total charge is, $q_1+q_2=-60\mu \text{C}$

$\Rightarrow q_1+2q_1=-60$

$\Rightarrow q_1=-20\mu \text{C}$, and

$q_2=-40\mu \text{C}$

Now,

Initial energy of the system is,

$U_i=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}\times \frac{(60\times {10}^{-6}{)}^2}{6\times {10}^{-6}}=0.3\text{mJ}$

Final energy of the system is,

$U_f=\frac{1}{2}[\frac{(20\times {10}^{-6}{)}^2}{3\times {10}^{-6}}+\frac{(40\times {10}^{-6}{)}^2}{6\times {10}^{-6}}]=0.2\text{mJ}$

Energy dissipated across the resistor is,

$U_i-U_f=0.3-0.2=0.1\text{mJ}$