Question

Two large conducting plates are placed parallel to each other with a separation d. An electron (-e, m) starting from rest near one of the plates reaches the other plate in time $t_0$. Find the surface charge density on the inner surface.

• A. $\frac{m{\epsilon }_{0}d}{e{t}_0^2}$
• B. $\frac{2m{\epsilon }_{0}d}{e{t}_0^2}$
• C. $\frac{m{\epsilon }_{0}d}{2e{t}_0^2}$
• D. $\frac{m{\epsilon }_{0}d}{4e{t}_0^2}$

Answer 1

The correct option is B $\frac{2m{\epsilon }_{0}d}{e{t}_0^2}$

The electric field between the plates is uniform and is

$E=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}$

Force on electron $F=eE=\frac{e\sigma }{{\epsilon }_0}$

Acceleration of electron $a=\frac{F}{m}=\frac{e\sigma }{{\epsilon }_{0}m}$

The electron moves from $\left(2\right)$ to $\left(1\right)$
$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow d=0+\frac{1}{2}\frac{e\sigma }{m{\epsilon }_0}{t}_0^2$

$\Rightarrow \sigma =\frac{2m{\epsilon }_{0}d}{e{t}_0^2}$

Hence, option (b) is the correct answer.