Question

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. $2F$

Answer 1

Answer: (B)

$F/2$

Answer 2

Initially F = qE and $E = \frac{\sigma}{\varepsilon_{0}}$ ∴ $F = \frac{q\sigma}{\varepsilon_{0}}$

If one plate is removed, then E becomes

$\frac{\sigma}{2\varepsilon_{0}}$

So $F' = \frac{q\sigma}{2\varepsilon_{0}} = \frac{F}{2}$