Question

Two thin concentric hollow conducting spheres of radii $R_1$ and $R_2$ carry charges $Q_1$ and $Q_2$ respectively. If $R_1<R_2$ then the potential at a point distant $r$ such that $R_1<r<R_2$ is

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{Q_1+Q_2}{r}$
• B. $\frac{1}{4\pi {\epsilon }_0}[\frac{Q_1}{r}+\frac{Q_2}{R_2}]$
• C. $\frac{1}{4\pi {\epsilon }_0}[\frac{Q_1}{R_1}+\frac{Q_2}{R_2}]$
• D. $\frac{1}{4\pi {\epsilon }_0}[\frac{Q_1}{R_1}+\frac{Q_2}{r}]$

Answer 1

The correct option is B $\frac{1}{4\pi {\epsilon }_0}[\frac{Q_1}{r}+\frac{Q_2}{R_2}]$

Sphere - 1

Charge $=Q_1$

Radius $=R_1$

Sphere - 2

Charge $=Q_2$

Radius $=R_2$

So, Potential due to sphere 2, can be determined as, $U_2=\frac{K{Q}_2}{R_2}$

Because potential inside a charged spherical conductor shell as same as that of potential on the surface of the shell.

for potential due to Sphere - 1, $U_1=\frac{{KQ}_1}{r}$

where $r$ is the distance from the center of the first sphere - 1, because in this case the charge is considered at the center of the sphere.

$\therefore$ The resultant potential will be given by,

$U=U_1+U_2$

$=\frac{{KQ}_1}{r}+\frac{{KQ}_2}{R_2}$

$=K[\frac{Q_1}{r}+\frac{Q_2}{R_2}]$

$\Rightarrow$ Option (B) is the correct option.