Question

A large flat metal surface has a uniform charge density, sigma. An electron of mass m and charge e leaves the surface at a point A with speed u , returns at B. The maximum value of AB is

Answer 1

From motion along X-axis in journey from A to B displacement in x direction is,=0

So, From 2nd equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

$0=ucos\theta T+\frac{1}{2}(-a)T^2⟹T=\frac{2ucos\theta }{a}.......\left(1\right)$

Field due to falt metal surface, $E=\frac{\sigma }{{ϵ}_o}$

Force on electron,$F=eE$

So acceleration will be along X-axis,$a=\frac{eE}{m}$

putting value of acceleration and electric feield in eq(1)

$T=\frac{2um{\epsilon }_{o}cos\theta }{e\sigma }$

So distance covered AB will be,

$=usin\theta \theta T=usin\theta \times \frac{2m{\epsilon }_{o}cos\theta }{e\sigma }$

$=\frac{m{\epsilon }_o{u}^{2}2sin\theta cos\theta }{e\sigma }=\frac{m{\epsilon }_o{u}^{2}sin2\theta }{e\sigma }$

$\theta ={45}^o$

$\frac{m{\epsilon }_o{u}^{2}sin2\theta }{e\sigma }=\frac{m{\epsilon }_o{u}^2}{e\sigma }$

So, distance covered AB will be,

$=\frac{m{\epsilon }_o{u}^2}{e\sigma }$