Question 196 of 311

Concentric metallic hollow spheres of radii R and 4R hold charges ${Q}_{1}$ and ${Q}_{2}$ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference $V\left(R\right)-V(4R)$ is:

A$\dfrac{{Q}_{2}}{4\pi {\epsilon }_{0}R}$B$\dfrac{3{Q}_{2}}{4\pi {\epsilon }_{0}R}$C$\dfrac{3{Q}_{1}}{16\pi {\epsilon }_{0}R}$D$\dfrac{3{Q}_{1}}{4\pi {\epsilon }_{0}R}$

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Answer: (C) $\dfrac{3{Q}_{1}}{16\pi {\epsilon }_{0}R}$

As given the surface charge density for both the spheres is same, hence
${\sigma }_{1}={\sigma }_{2} ∴\dfrac{{Q}_{1}}{{A}_{1}}=\dfrac{{Q}_{2}}{{A}_{2}}\Rightarrow \dfrac{{Q}_{1}}{4\pi {R}^{2}}=\dfrac{{Q}_{2}}{4\pi {\left(4R\right)}^{2}} ∴{Q}_{2}=16{Q}_{1}$
From above diagram, the electric potential for both the spheres is
${V}_{Inner}=\dfrac{{KQ}_{1}}{R}+\dfrac{{KQ}_{2}}{4R}=\dfrac{K}{R}\left({Q}_{1}+\dfrac{16{Q}_{1}}{4}\right)=\dfrac{5{KQ}_{1}}{R}.$
In the same way for outer sphere,
${V}_{Outer}=\dfrac{{KQ}_{1}}{4R}+\dfrac{{KQ}_{2}}{4R}=\dfrac{K}{4R}\left(16{Q}_{1}+{Q}_{1}\right)=\dfrac{17{KQ}_{1}}{4R}$.
Now the potential difference is,
$∆V={V}_{inner}-{V}_{outer}$
$=\dfrac{5{KQ}_{1}}{R}-\dfrac{17{KQ}_{1}}{4R}=\dfrac{3{Q}_{1}}{16{\pi \epsilon }_{0}R}$.