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JEE Main 2022 (27 Jun, Shift 1) · Parallel Plate Capacitor

Parallel Plate Capacitor question from JEE Main 2022 (27 Jun, Shift 1)

Q11JEE Main 2022 · 27 JUN · SHIFT 1Medium

Parallel Plate Capacitor

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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Answer: (a)

Between the plates each plate gives $\frac{\sigma}{2\varepsilon_0}$ ; together $E = \frac{\sigma}{\varepsilon_0}$ .
With one plate removed, $E' = \frac{\sigma}{2\varepsilon_0} = \frac{E}{2}$ .
$F' = \frac{F}{2} = \frac{10}{2}$
So the answer is (a) $5\text{ N}$ .

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