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, 14πϵok3q2a2=14πϵo3q2a2−F2 which gives us F2=3q24πϵoa2(1−1k) By putting k = 2 and comparing, we get n = 1.5