Q.
The electron is projected from a distance d and with initial velocity u parallel to a uniformly charged flat conducting plate as shown in figure. It strikes the plate after travelling a distance $l$ along the direction. The surface charge density of conducting plate is equal to
A) $\frac{2d\varepsilon_{0}mu^{2}}{el^{2}}$
B) $\frac{2d\varepsilon_{0}mu^{\ }}{el^{\ }}$
C) $\frac{d\varepsilon_{0}mu^{2}}{el^{\ }}$
D) $\frac{d\varepsilon_{0}mu^{\ }}{el^{\ }}$
Explanation
(a) Here, $E = \frac{\sigma}{\varepsilon_{0}}\ and\ t = \frac{1}{u}$
Along Y-axis, $\ u = 0,\ a = \frac{f}{m} = \frac{eE}{m}$
$s = d = \frac{1}{2}at^{2} = \frac{1}{2}\frac{eE}{m}t^{2} = \frac{1}{2}\frac{e\sigma}{m\varepsilon_{0\ }}\frac{l^{2}}{u^{2}}$
$\sigma = \frac{2d\varepsilon_{0}mu^{2}}{el^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.