Question:
The given graph shows variation (with distance \(r\) form centre) of :
The given graph shows variation (with distance \(r\) form centre) of :
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The quantity is constant for \(r < r_0\) and falls as \(\frac{1}{r}\) outside, continuous at \(r_0\): this is the potential of a charged spherical shell.
Updated On: Nov 5, 2026
- Electric field of a uniformly charged sphere
- Electric field of a uniformly charged spherical shell
- Potential of a uniformly charged sphere
- Potential of a uniformly charged spherical shell
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The Correct Option is D
Solution and Explanation
For a uniformly charged spherical shell of radius \(r_0\) and charge \(Q\):
Electric field: \(E = 0\) for \(r < r_0\) and \(E = \frac{kQ}{r^2}\) for \(r > r_0\), i.e. it jumps from zero to \(\frac{kQ}{r_0^2}\) at \(r_0\). The graph shown is non-zero and flat inside and continuous, so it is not the field of a shell.
Potential: \(V = \frac{kQ}{r_0}\) (constant) for \(r \le r_0\) and \(V = \frac{kQ}{r}\) (decreasing hyperbolically) for \(r \ge r_0\), continuous at \(r_0\). This is the graph shown.
For a uniformly charged solid sphere, the field rises linearly inside and the potential is not constant inside, so options (1) and (3) do not fit either.
Answer: (4) Potential of a uniformly charged spherical shell
Electric field: \(E = 0\) for \(r < r_0\) and \(E = \frac{kQ}{r^2}\) for \(r > r_0\), i.e. it jumps from zero to \(\frac{kQ}{r_0^2}\) at \(r_0\). The graph shown is non-zero and flat inside and continuous, so it is not the field of a shell.
Potential: \(V = \frac{kQ}{r_0}\) (constant) for \(r \le r_0\) and \(V = \frac{kQ}{r}\) (decreasing hyperbolically) for \(r \ge r_0\), continuous at \(r_0\). This is the graph shown.
For a uniformly charged solid sphere, the field rises linearly inside and the potential is not constant inside, so options (1) and (3) do not fit either.
Answer: (4) Potential of a uniformly charged spherical shell
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