Solution: Potential at B due to shell A = kQ2aPotential at B due to shell B = 2kQ2aPotential at B due to shell C = -kQ3aSo Potential at B =kQ2a+2kQ2a+-kQ3a=7kQ6aso n=7

Find the value of n

Solution:

P o t e n t i a l a t B d u e t o s h e l l A = \frac{k Q}{2 a} P o t e n t i a l a t B d u e t o s h e l l B = \frac{2 k Q}{2 a} P o t e n t i a l a t B d u e t o s h e l l C = \frac{- k Q}{3 a} S o P o t e n t i a l a t B = \frac{k Q}{2 a} + \frac{2 k Q}{2 a} + \frac{- k Q}{3 a} = \frac{7 k Q}{6 a} s o n = 7

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