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Q.

Two charged particles A and B are kept at a certain separation and magnitude of electric force acting on each particle is F. Now additional 50% of charge of A is added to A and the separation between A and B is halved. Now what will be the magnitude of electric force acting on each particle?

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a

4F3

b

6 F

c

4 F

d

8F3

answer is D.

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Detailed Solution

detailed_solution_thumbnail

F=KQ1Q2r2 Here Q1=charge on A and Q2= charge on B

Q11=Q1+50% of Q1=32Q1

∴ F1=KQ11.Q2r22=K3Q12.Q2r24=6KQ1Q2r2

⇒F1=6F

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