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Question

Two point charges placed at a distance r in air exert a force F on each other. What will be the distance at which they will experience force 4F, when placed in a medium of dielectric constant 16?

A
r
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B
r4
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C
r8
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D
2r
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Solution

The correct option is C r8
When charges are placed in air, the electric force between them is:

Fe=14πϵ0q1q2r2 .......(1)

However, when charge is placed in medium,

Fe=14πϵq1q2r2

Where ϵ=ϵ0×ϵr is called permittivity of medium.

So, when charge placed in medium,

4F=14πϵrϵ0q1q2R2 (2)

From eqs.(1) and (2),

4=r2ϵrR2 (ϵr=16)

R2=r264

R=r8

Hence, option (c) is the correct answer.

Why this question ?Bottom line: When charges are placed at same separation (r)in a medium of dielectric constant ϵr the force betweenthem decreases.i.e. F=Fϵr

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