• question_answer There are two hollow concentric spheres with radii ${{R}_{1}}$and R for the inner and outer spheres respectively. The inner sphere is given a charge ${{Q}_{1}}$and the outer a charge${{Q}_{2}}.$Then the potential at the surface of the inner sphere is A)  $\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{R}_{1}}}+\frac{{{Q}_{2}}}{{{R}_{2}}} \right)$B) $\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{R}_{1}}}-\frac{{{Q}_{2}}}{{{R}_{2}}} \right)$C) $\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{R}_{1}}} \right)$                      D)  0

Potential at the surface of the inner sphere due to the charges of the outer sphere $={{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{2}}}{{{R}_{2}}}$ Potential at the surface of the inner sphere due to its own charges ${{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}}{{{R}_{1}}}$ .'. Net potential on the surface of the inner sphere $=V={{V}_{1}}+{{V}_{2}}$ or,        $V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{R}_{1}}}+\frac{{{Q}_{2}}}{{{R}_{2}}} \right)$ Hence, the correction option is [a].