Identify the graph qualitatively

For this standard electrostatics question, the graph implied is the one where the quantity is:

• constant from $r=0$ to $r=R$ (inside the object), and
• then decreases as $\dfrac{1}{r}$ for $r>R$ outside.

We compare this behavior with the known formulas for electric field and potential of a uniformly charged sphere and spherical shell.

Check each option

Option A: Electric field of a uniformly charged sphere

For a solid uniformly charged sphere, $E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Qr}{R^3}, \quad r<R$ and $E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}, \quad r\ge R$

So inside, $E \propto r$ (starts from zero and increases linearly), not constant. Hence A is not correct.

Option B: Electric field of a uniformly charged spherical shell

For a spherical shell, $E(r)=0, \quad r<R$ and $E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}, \quad r\ge R$

Inside it is zero, outside it falls as $1/r^2$. This does not match a graph that is constant inside and falls as $1/r$ outside. Hence B is not correct.

Option C: Potential of a uniformly charged sphere

For a uniformly charged solid sphere, $V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{2R}\left(3-\frac{r^2}{R^2}\right), \quad r<R$ and $V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}, \quad r\ge R$

Inside, potential is not constant; it varies parabolically with $r$. Hence C is not correct.

Option D: Potential of a uniformly charged spherical shell

For a uniformly charged spherical shell, $V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}, \quad r\le R$ and $V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}, \quad r\ge R$

So:

• inside the shell, potential is constant,
• outside, it decreases as $1/r$.

This matches the described graph exactly.

Final answer

Therefore, the graph represents the potential of a uniformly charged spherical shell.

$\boxed{\text{D}}$

Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

So, they agree.