1. Electric potential of a charged spherical conductor

For a spherical conductor of radius $R$ carrying charge $Q$:

• Outside the conductor $(r \ge R)$, it behaves like a point charge: $V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$

• On the surface $(r=R)$: $V(R)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$

• Inside the conductor $(r < R)$, the electric field is zero, so the potential remains constant and equal to the surface potential: $V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$

2. Nature of the graph

So the variation of $V$ with $r$ is:

• From $r=0$ to $r=R$: constant
• For $r>R$: decreases as $\dfrac{1}{r}$

Thus the correct graph must show:

• a horizontal line from $0$ to $R$
• then a falling curve like $1/r$ beyond $R$

3. Identify the correct option

Among the given choices, this corresponds to Option D.

4. Comparison with stored answer

Stored correct answer: D

Derived answer: D

Hence, the derived answer agrees with the stored correct answer.