Q129 / 180 · KCET 2013 · Physics

The concentric spheres of radii $R$ and $r$ have positive charges $q_{1}$ and $q_{2}$ with equal surface charge densities. What is the electric potential at their common centre?

  • Correct answer
    A
    $\frac{\sigma}{\varepsilon_{0}}(R+r)$
  • B
    $\frac{\sigma}{\varepsilon_{0}}(R-r)$
  • C
    $\frac{\sigma}{\varepsilon_{0}}\left(\frac{1}{R}+\frac{1}{r}\right)$
  • D
    $\frac{\sigma}{\varepsilon_{0}}\left(\frac{1}{R}\right)$

Solution

Potential at the centre due to sphere with radius $R$, $V_{1}=\frac{q_{1}}{4 \pi \varepsilon_{0} R}$ Potential at the centre due to sphere with radius $r$, $V_{2}=\frac{q_{2}}{4 \pi \varepsilon_{0} r}$ Given, surface charge densities for both spheres $\sigma=\sigma_{R}=\sigma_{r}$ i.e., $\begin{aligned} \text { i.e., } & \frac{q_{1}}{4 \pi \varepsilon_{0} R^{2}} &=\frac{q_{2}}{4 \pi \varepsilon_{0} r^{2}} \\ \Rightarrow & q_{1} &=\frac{R^{2}}{r^{2}} q_{2} \end{aligned}$ Total potential at the centre due to both spheres $\begin{aligned} V &=V_{1}+V_{2} \\ &=\frac{q_{1}}{4 \pi \varepsilon_{0} R}+\frac{q_{2}}{4 \pi \varepsilon_{0} r}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}}{R}+\frac{q_{2}}{r}\right) \\ &=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{R^{2}}{r^{2}} \frac{q_{2}}{R}\right)+\frac{q_{2}}{r}=\frac{q_{2}}{4 \pi \varepsilon_{0}}\left(\frac{R+r}{r^{2}}\right) \\ &=\frac{q_{2}}{4 \pi r^{2} \varepsilon_{0}}(R+r)=\frac{\sigma}{\varepsilon_{0}}(R+r) \end{aligned}$

Hence, the correct answer is option: 'A'.

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