Question

Force acting on a test charge between the plates of a parallel capacitor is $F$. If one of the plates is removed, then the force on the same test charge will be:

• A. Zero
• B. $F$
• C. $2F$
• D. $\frac{F}{2}$

Answer 1

The correct option is D $\frac{F}{2}$

The electric between the plate is $E=\frac{\sigma }{{ϵ}_0}$ where $\sigma =$ surface charge density of the plates.
The force on test charge q is $F=qE$.
When one plate removed, the field becomes $E^{\prime }=\frac{\sigma }{2{ϵ}_0}=\frac{E}{2}$
Now the force , $F^{\prime }=q{E}^{\prime }=\frac{qE}{2}=\frac{F}{2}$