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Question

Two point charges of magnitude q and 3q are located at a distance 'a' in dielectric medium of dielectric constant 2. The force applied by the larger charge 3q on dielectric is nq24πϵoa2. Here n is

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Solution

Fnet on 3q is :
Fnet=F1+F2
where F1 is Force on 3q due to q
F2 is force on 3q due to dielectric

So,
14πϵok 3q2a2=14πϵo3q2a2F2
which gives us
F2=3q24πϵoa2 (11k)
By putting k = 2 and comparing, we get n = 1.5

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