Question

A charge "q" is released from rest from a height "h" above a charged plate. The charged plate creates a uniform electric field "E" between the charge and the plate. Determine the velocity, the charge will have just before it strikes the plate that it is attracted to :

(Ignore any effect of gravity)

• A. $v=\sqrt{\frac{2m}{qEh}}$
• B. $v=\sqrt{\frac{m}{2qEh}}$
• C. $v=\sqrt{\frac{qEh}{2m}}$
• D. $v=\sqrt{\frac{2qEh}{m}}$

Answer 1

The correct option is D $v=\sqrt{\frac{2qEh}{m}}$

The constant force acting on the charge is $F=qE$.
The acceleration of charge due to this force=$\frac{F}{m}=\frac{qE}{m}=a$

From the equation of motion,

$v^2-u^2=2as$

$⟹v^2=2\times \frac{qE}{m}\times h$

$⟹v=\sqrt{\frac{2qEh}{m}}$

Hence correct answer is option D.