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Solution :

Radius of the outer shell =`r_(1)` <br> radius of the inner shell =`r_(2)` <br> The inner surface of the outer shell has charge +Q <br> the outer surface of the inner shell induced charge -Q. <br> potential difference between the two shells is given by, <br> `V=Q/(4piin_(0)r_(2))-Q/(4piin_(0)r_(1))` <br> where, <br> `in_(0)`=permitivity of free space <br> `V=Q/(4pi in_(0))[1/(r_(2))-1/(r_(1))]` <br> `=(Q(r_(1)-r_(2)))/(4piin_(0)r_(1)r_(2))` <br> capacitance of the given system is given by <br> `C=("Charge (Q)")/("Potential difference (V)")` <br> `=(4pi in_(0)r_(1)r_(2))/(r_(1)-r_(2))` <br> Henced proved.Related Video

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