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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 :

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a

20N

b

zero

c

10N

d

5N

answer is A.

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Detailed Solution

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Question Image

$F = q E = q (\frac{Q}{A ϵ_{0}}) = \frac{q Q}{A ϵ_{0}} = 10 N$

Now, when one plate is removed.

Question Image

$E^{'} = \frac{Q}{2 A ϵ_{0}} F = q E^{'} = \frac{Q q}{2 A ϵ_{0}} = 5 N$

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