JEE Main & Advanced Physics Electrostatics & Capacitance Potential Due to Concentric Spheres 

(1) If two concentric conducting shells of radii \[{{r}_{1}}\] and \[{{r}_{2}}({{r}_{2}}>{{r}_{1}})\] carrying uniformly distributed charges \[{{Q}_{1}}\] and \[{{Q}_{2}}\] respectively. Potential at the surface of each shell

\[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{Q}_{1}}}{{{r}_{1}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{Q}_{2}}}{{{r}_{2}}}\]

\[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{Q}_{1}}}{{{r}_{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{Q}_{2}}}{{{r}_{2}}}\]

(2) The figure shows three conducting concentric shell of radii \[a,\,\,b\] and \[c(a<b<c)\] having charges \[{{Q}_{a}},\,{{Q}_{b}}\] and \[{{Q}_{c}}\] respectively

Potential at A;

\[{{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{Q}_{a}}}{a}+\frac{{{Q}_{b}}}{b}+\frac{{{Q}_{c}}}{c} \right]\]

Potential at B;

\[{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{Q}_{a}}}{b}+\frac{{{Q}_{b}}}{b}+\frac{{{Q}_{c}}}{c} \right]\]

Potential at C;

\[{{V}_{C}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{Q}_{a}}}{c}+\frac{{{Q}_{b}}}{c}+\frac{{{Q}_{c}}}{c} \right]\]

(3) The figure shows two concentric spheres having radii \[{{r}_{1}}\] and \[{{r}_{2}}\] respectively \[({{r}_{2}}>{{r}_{1}})\]. If charge on inner sphere is +Q and outer sphere is earthed then

(i) Potential at the surface of outer sphere

\[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{{{r}_{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q'}{{{r}_{2}}}=0\]

\[\Rightarrow \] \[Q'=-Q\]

(ii) Potential of the inner sphere

\[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{{{r}_{1}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{(-Q)}{{{r}_{2}}}\]\[=\frac{Q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]\]

(4) In the above case if outer sphere is given a charge +Q and inner sphere is earthed then

(i) In this case potential at the surface of inner sphere is zero, so if \[Q'\] is the charge induced on inner sphere

then \[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{Q'}{{{r}_{1}}}+\frac{Q}{{{r}_{2}}} \right]=0\]

i.e.,   \[Q'=-\frac{{{r}_{1}}}{{{r}_{2}}}Q\]

(Charge on inner sphere is less than that of the outer sphere.)

(ii) Potential at the surface of outer sphere        

\[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q'}{{{r}_{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{{{r}_{2}}}\]        

\[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}{{r}_{2}}}\left[ -Q\frac{{{r}_{1}}}{{{r}_{2}}}+Q \right]\] \[=\frac{Q}{4\pi {{\varepsilon }_{0}}{{r}_{2}}}\left[ 1-\frac{{{r}_{1}}}{{{r}_{2}}} \right]\]