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  4. A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be :

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Q. A force of $10\, N$ acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be :

JEE MainJEE Main 2022Electrostatic Potential and Capacitance

Solution:

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$F=q E=q\left(\frac{Q}{A \in_{0}}\right)=\frac{q Q}{A \in_{0}}=10\, N$
Now, when one plate is removed.
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$E^{\prime}=\frac{Q}{2 A \epsilon_{0}}$
$F=q E^{\prime}=\frac{Q q}{2 A \epsilon_{0}}=5 \,N$