Question

An electron moves straight inside a charged parallel plate capacitor of uniform surface charge density $\sigma$. The space between the plates is filled with constant magnetic field of induction $\overrightarrow{B}$. The time of straight line motion of the electron in the capacitor is:

• A. $\frac{e\sigma }{{ϵ}_{0}lB}$
• B. $\frac{{ϵ}_{0}lB}{\sigma }$
• C. $\frac{e\sigma }{{ϵ}_{0}B}$
• D. $\frac{{ϵ}_{0}B}{e\sigma }$

Answer 1

The correct option is B $\frac{{ϵ}_{0}lB}{\sigma }$

The net electric field
$E={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$
$\Rightarrow =\frac{\sigma }{2{ϵ}_0}+\frac{\sigma }{2{ϵ}_0}=\frac{\sigma }{2{ϵ}_0}$
The net force acting on the electron is zero because it moves with constant velocity, due to its motion on straight line.
$\Rightarrow {\overrightarrow{F}}_{net}={\overrightarrow{F}}_e+{\overrightarrow{F}}_m=0$
$|{\overrightarrow{F}}_e|=|{\overrightarrow{F}}_m|$
$\Rightarrow eE=eVB\Rightarrow V=\frac{E}{B}=\frac{\sigma }{{ϵ}_{0}B}$
$\therefore$ The time of motion in side the capacitor $=t=\frac{l}{V}=\frac{ϵlB}{\sigma }$.