Question

A capacitor of capacitance $100\mu \text{F}$ is connected across a battery of e.m.f $6.0\text{V}$ through a resistance of $20\text{k}\Omega$ for $4.0\text{s}$. The battery is then replaced by a thick wire, what will be the charge on the capacitor $4.0\text{s}$ after the battery is disconnected ?

• A. $70\mu \text{C}$
• B. $80\mu \text{C}$
• C. $60\mu \text{C}$
• D. $40\mu \text{C}$

Answer 1

The correct option is A $70\mu \text{C}$

Initially charging:


We know that, $$q=q_0[1-e^{\frac{-t}{RC}}]$$
$\Rightarrow q=V_{0}C[1-e^{\frac{-t}{RC}}]$
Substituting the values, at $t=4\text{s}$,

$q=(6\times 100)[1-e^{-\frac{4}{(20\times {10}^3)\times (100\times {10}^{-6})}}]\mu \text{C}$

$\Rightarrow q=5.19\times {10}^{-4}\text{C}$

Discharging: when battery is replaced


$q^{\prime }=q{e}^{\frac{-t}{RC}}$

Where, $q$ is the charge just before discharging started. Substituting the values,

$q^{\prime }=5.19\times {10}^{-4}\times \left[e^{-2}\right]=70\mu \text{C}$

Hence, option (a) is the correct answer.