Question

Two concentric, thin metallic spheres of radii $R_1$ And $R_2$($R_1>R_2$) bear charges $Q_1$and $Q_2$respectively. Calculate the potential at a distance $r$ between $R_1$ and $R_2$.

Answer 1

Step 1: Given data

Two metallic spheres having raddii $R_1$ and $R_2$($R_1>R_2$) containing charge $Q_1$ and $Q_2$ respectively.

Step 2: Calculating potential due to $Q_1$

Potential due to a charged sphere of radius $R_1$ containing charge $Q_1$ at a distance $r$ is given by

$V_1=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1}{R_1}........\left(1\right)$

As $r$ remains inside the sphere, the potential will be constant.

Step 3: Calculating potential due to $Q_2$

Now the potential due to a charged sphere of radius $R_2$ containing a charge $Q_2$,at a distance $r$ is given by

$V_2=\frac{1}{4\pi {\epsilon }_0}\frac{Q_2}{r}.........\left(2\right)$

As $r$ is outside the sphere.

Step 4: Calculating total potential

Now the total potential at a distance $r$ between $R_1$ and $R_2$can be given by,

$\begin{array}{rcl}{V}_r& =& V_1+V_2\\ & =& \frac{1}{4\pi {\epsilon }_0}\frac{Q_1}{R_1}+\frac{1}{4\pi {\epsilon }_0}\frac{Q_2}{r}\\ & =& \frac{1}{4\pi {\epsilon }_0}(\frac{Q_1}{R_1}+\frac{Q_2}{r})\end{array}$

Hence, the potential at a distance $R$ between $R_1$ and $R_2$ is $\begin{array}{rcl}{v}_r& =& \frac{1}{4\pi {\epsilon }_0}(\frac{Q_1}{R_1}+\frac{Q_2}{r})\end{array}$.