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 ...... $N$
JEE MAIN 2022, Medium
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$F=q E=q\left(\frac{Q}{A \epsilon_{0}}\right)=\frac{q Q}{A \epsilon_{0}}=10 \; N$
Now, when one plate is removed.
$E^{\prime}=\frac{Q}{2 A \epsilon_{0}}$
$F=q E^{\prime}=\frac{Q q}{2 A \epsilon_{0}}=5 \; N$
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