Q. Force acting upon a charged particle kept between the plates of a charged condenser is $F.$ If one plate of the condenser is removed, then the force acting on the same particle will become
A) 0
B) $F/2$
C) $F$
D) $2F$
Explanation
(b) Initially $F = qE$ and $E = \frac{\sigma}{\varepsilon_{0}}\ \therefore F = \frac{q\sigma}{\varepsilon_{0}}$
If one plate is removed, then $E$ becomes $\frac{\sigma}{2\varepsilon_{0}}$
So $F^{'} = \frac{q\sigma}{2\varepsilon_{0}} = \frac{F}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.