Question

The switch is closed at $t=0$, when capacitor was completely uncharged. Which of the options are correct for the charging process ?

• A. Initial current through ${}^{\prime }{R}^{\prime }$ is $\frac{E}{R}$
• B. Heat produced during charging is $\frac{1}{2}C{\epsilon }^2$
  
• C. Attractive force between plates of Area A is $\left[\frac{C^2{\epsilon }^2}{2{ϵ}_{0}A}\right]$ in steady state
• D. Maximum rate possible at which energy is being stored in capacitor is $\frac{{\epsilon }^2}{4R}$

Answer 1

Answer: (A), (B), (C), (D)

Maximum rate possible at which energy is being stored in capacitor is $\frac{{\epsilon }^2}{4R}$

Just after closing the switch potential difference across capacitor is zero, so $I_{Initial}=\frac{\epsilon }{R}$
From conservation of energy
$W_{cell}=\Delta H+\Delta U$
$\left(c\epsilon \right)\epsilon =\Delta H+\frac{1}{2}C{ϵ}^2$
Heat produced during charging process is
$\Rightarrow \Delta H=\frac{1}{2}C{ϵ}^2$
Charge on capacitor is
$Q=Cϵ\text{Force between plates of capacitor}=\left(\frac{Q/A}{2{ϵ}_0}\right)\times Q$
$F=\frac{C^2{ϵ}^2}{2A{ϵ}_0}$

Equating input power with potput power
$\Rightarrow \epsilon I=\text{Rate of energy storage}-I^{2}R$
`Rate of energy storing $P=\epsilon I-I^{2}R$
for max $P\Rightarrow \frac{dP}{dI}=0$
$\Rightarrow \epsilon -2IR=0$
$\Rightarrow I=\frac{\epsilon }{2R}$
$\Rightarrow p_{max}=\epsilon \left(\frac{\epsilon }{2R}\right)-{\left(\frac{\epsilon }{2R}\right)}^{2}R=\frac{{\epsilon }^2}{4R}$