2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Electrostatics & Capacitance

JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Self Evaluation Test - Electrostatic Potential and Capacitance

  • question_answer
    A charge Q is distributed over two concentric hollow spheres of radii r and R(R>r) such that the surface densities are equal. The potential at the common center is \[\frac{1}{4\pi {{\varepsilon }_{0}}}\]times-

    A) \[Q\left[ \frac{r+R}{{{r}^{2}}+{{R}^{2}}} \right]\]                 

    B) \[\frac{Q}{2}\left[ \frac{r+R}{{{r}^{2}}+{{R}^{2}}} \right]\]

    C) \[2Q\left[ \frac{r+R}{{{r}^{2}}+{{R}^{2}}} \right]\]   

    D) zero

    Correct Answer: A

    Solution :

    [a] \[{{q}_{r}}+{{q}_{R}}=Q\]                                    ?.(1) \[\frac{{{q}_{r}}}{{{q}_{R}}}=\frac{4\pi {{r}^{2}}\sigma }{4\pi {{R}^{2}}\sigma }=\frac{{{r}^{2}}}{{{R}^{2}}}\Rightarrow \frac{{{q}_{r}}}{{{q}_{R}}}=\frac{{{r}^{2}}}{{{R}^{2}}}\]               ?.(2) From (1) and (2) \[{{q}_{r}}=\frac{Q{{r}^{2}}}{{{R}^{2}}+{{r}^{2}}}\]and \[{{q}_{R}}=\frac{Q{{R}^{2}}}{{{R}^{2}}+{{r}^{2}}}\] So, \[V=\frac{{{q}_{r}}}{4\pi {{\varepsilon }_{0}}r}+\frac{{{q}_{R}}}{4\pi {{\varepsilon }_{0}}R}\] \[\Rightarrow V=\frac{Q}{4\pi {{\varepsilon }_{0}}}\left( \frac{r+R}{{{r}^{2}}+{{R}^{2}}} \right)\]


You need to login to perform this action.
You will be redirected in 3 sec spinner