Three concentric, thin, spherical, metallic shells have radii $1,2 ,$ and $4cm$ and they are held at potentials 10,0 and $40V$ respectively. Taking the origin at the common centre, calculate the :

Potential at $r=1.25cm$

$\text{ Let }{q}_1:q_2\text{ and }{q}_3\text{ be the respective charges. Then, }$

$$\begin{gathered} 10=\frac{9\times {10}^9}{{10}^{-2}}\left[\frac{q_1}{1}+\frac{q_2}{2}+\frac{q_3}{4}\right] \\ 0=\frac{9\times {10}^9}{{10}^{-2}}\left[\frac{q_1}{2}+\frac{q_2}{2}+\frac{q_3}{4}\right] \\ \text{ and }40=\frac{9\times {10}^9}{{10}^{-2}}\left[\frac{q_1}{4}+\frac{q_2}{4}+\frac{q_3}{4}\right] \end{gathered}$$

Solving these equations, we get

$$\begin{gathered} q_1=+\frac{200}{9}\times {10}^{-12}C, q_2=-200\times {10}^{-12}C\text{ and } \\ {q}_3=\frac{3200}{9}\times {10}^{-12}C \end{gathered}$$

At $r=1.25cm$

$$\begin{gathered} V=\frac{9\times {10}^9}{{10}^{-2}}\left[\frac{\left(200/9\right)\times {10}^{-12}}{1.25}-\frac{200\times {10}^{-12}}{2}+\frac{(32009\times {10}^{-12})}{4}\right] \\ =6 V \end{gathered}$$