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
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(b) 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 $F' = \frac{{q\sigma }}{{2{\varepsilon _0}}} = \frac{F}{2}$
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