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Q.

The force of 10N 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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a

20 N

b

5 N

c

10

d

Zero

answer is A.

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Detailed Solution

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E between two plates $\frac{σ}{ε_{0}}$ and due to one plate is $\frac{σ}{2 ε_{0}}$ so the force will be halved so new force F = 5N

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