Question

The electron is projected from a distance $d$ and with an initial velocity $v_0$ parallel to a uniformly charged flat conducting plate as shown. It strikes the plate after travelling a distance $l$ along the direction. The surface charge density of the conducting plate is equal to

• A. $\frac{2d{ϵ}_{0}m{v}_0}{el}$
• B. $\frac{d{ϵ}_{0}m{v_0}^2}{el}$
• C. $\frac{d{ϵ}_{0}m{v}_0}{el}$
• D. $\frac{2d{ϵ}_{0}m{v_0}^2}{e{l}^2}$

Answer 1

The correct option is D $\frac{2d{ϵ}_{0}m{v_0}^2}{e{l}^2}$

Let field along Y-direction be E.

And, there is no field in the X-direction.

Time of movement$=t=\frac{l}{v_x}$

$⟹t=\frac{l}{v}$

Now, let acceleration of electron in Y-direction is a and initial speed is 0 along Y-axis.

$d=\frac{1}{2}a{t}^2$

$⟹d=\frac{a{l}^2}{2v^2}$

$⟹a=\frac{2d{v}^2}{l^2}$

Now, field near plate$=E=\frac{\sigma }{{ϵ}_o}$

And, hence force on electron$=F=eE=\frac{e\sigma }{{ϵ}_o}$

And, acceleration$=a=\frac{F}{m}=\frac{e\sigma }{m{ϵ}_o}$

Hence, $a=\frac{e\sigma }{m{ϵ}_o}=\frac{2d{v}^2}{l^2}$

$⟹\sigma =\frac{2d{ϵ}_{o}m{v}^2}{e{l}^2}$

Answer-(D)