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 $4.4\mu \text{s}$ when the plates are connected by a thin wire. Find the resistance of the wire. $(\ln 3=1.0985)$

• A. $3.0\Omega$
• B. $2.0\Omega$
• C. $4.0\Omega$
• D. $1.0\Omega$

Answer 1

The correct option is B $2.0\Omega$

Given,

Initial electric field, $E_i=E$

Final electric field, $E_f=\frac{E}{3}$

If $\sigma$ is the charge density of the plates then the electric field between the plates will be,

$E=\frac{\sigma }{{\epsilon }_0}$

So, ratio of electric field will be

$\frac{E_i}{E_f}=\frac{{\sigma }_i/{\epsilon }_0}{{\sigma }_f/{\epsilon }_0}$

$\Rightarrow \frac{{\sigma }_i}{{\sigma }_f}=\frac{E}{E/3}=3$

$\therefore \frac{Q_i}{Q_f}=3[\because \sigma =\frac{Q}{A}]$

As per question, the given capacitor is discharging through a resistor. So, while discharging, the charge after time $t$ is,

$Q_f=Q_i^{\frac{-t}{RC}}$

Substituting the values,

$\Rightarrow \frac{Q_i}{3}=Q_i{e}^{\frac{-4.4}{2R}}$

$\Rightarrow 3=e^{\frac{2.2}{R}}$

$\Rightarrow R=\frac{2.2}{\ln 3}=2.0\Omega$

Hence, option $\left(B\right)$ is correct.