- Home
- Standard 12
- Physics
A charge $Q$ is distributed over two concentric conducting thin spherical shells radii $r$ and $R$ $( R > r ) .$ If the surface charge densities on the two shells are equal, the electric potential at the common centre is
$\frac{1}{4 \pi \varepsilon_{0}} \frac{( R +2 r ) Q }{2\left( R ^{2}+ r ^{2}\right)}$
$\frac{1}{4 \pi \varepsilon_{0}} \frac{( R + r )}{2\left( R ^{2}+ r ^{2}\right)} Q$
$\frac{1}{4 \pi \varepsilon_{0}} \frac{( R + r )}{\left( R ^{2}+ r ^{2}\right)} Q$
$\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 R+r)}{\left(R^{2}+r^{2}\right)} Q$
Solution
Let the charges on inner and outer spheres are $Q _{1}$ and $Q _{2}$
since charge density $'\sigma'$ is same for both spheres, so
$\sigma=\frac{Q_{1}}{4 \pi r^{2}}=\frac{Q_{2}}{4 \pi R^{2}} \Rightarrow \frac{Q_{1}}{Q_{2}}=\frac{r^{2}}{R^{2}}$
$Q _{1}+ Q _{2}= Q \Rightarrow \frac{ Q _{2} r ^{2}}{ R ^{2}}+ Q _{2}= Q$
$\Rightarrow Q_{2}=\frac{Q R^{2}}{\left(r^{2}+R^{2}\right)}$
$Q_{1}=\frac{r^{2}}{R^{2}} \cdot \frac{Q R^{2}}{\left(R^{2}+r^{2}\right)}=\frac{Q r^{2}}{\left(R^{2}+r^{2}\right)}$
Potential at centre ${ }^{\prime} O ^{\prime}=\frac{ k Q _{1}}{ r }+\frac{ k Q _{2}}{ R }$
$= k \left[\frac{ Qr ^{2}}{ r \left( R ^{2}+ r ^{2}\right)}+\frac{ Q R ^{2}}{ R \left( R ^{2}+ r ^{2}\right)}\right]$
$=\frac{ k Q ( r + R )}{\left( R ^{2}+ r ^{2}\right)}=\frac{1}{4 \pi \epsilon_{0}} \frac{( R + r )}{\left( R ^{2}+ r ^{2}\right)} Q$
