Question

The electric field between the plates of a parallel - plate capacitor of capacitance $2.0\mu \text{F}$ drops to one-third of its initial value in $4.4\text{ms}$ when the plates are connected by a thin wire. Find the resistance of the wire.

• A. $\frac{2200}{\ln 3}\Omega$
  
• B. $\frac{1100}{\ln 3}\Omega$
  
• C. $\frac{2200}{\ln 6}\Omega$
  
• D. $\frac{1100}{\ln 6}\Omega$
  
  

Answer 1

The correct option is A $\frac{2200}{\ln 3}\Omega$


As, $E=\frac{V}{d}$

$\therefore E\propto V$

$\because$ Charge on the capacitor is directly proportional to potential differnce between the plates.

$q=\frac{q_0}{3},$

In the case of discharging capacitor at any instant time $t$ is,
$$q=q_0{e}^{-t/\tau }$$

$\therefore \frac{q_0}{3}=q_0{e}^{-t/\tau }$

$3=e^{\frac{t}{\tau }}$

Taking log on both sides,

$\ln 3=\frac{t}{\tau }$

$\therefore t=\tau \ln 3=RC\ln 3$

$\therefore R=\frac{t}{C\ln 3}=\frac{4.4\times {10}^{-3}}{2\times {10}^{-6}\left(\ln 3\right)}$

$\Rightarrow R=\frac{2200}{\ln 3}\Omega$

Hence, option (a) is the correct answer.

why this question ? Key concept : Electric field between the plates of capacitor has direct relationship with charge on the capacitor