1. Electric field due to a uniformly charged insulating solid sphere

For a solid insulating sphere of radius $R$ with uniform volume charge density, the electric field varies differently in two regions:

• Inside the sphere: $r < R$
• Outside the sphere: $r \ge R$

2. Field inside the sphere $(r<R)$

Using Gauss’s law,

$\oint \vec E \cdot d\vec A = \frac{q_{\text{enc}}}{\varepsilon_0}$

For a Gaussian surface of radius $r<R$,

$q_{\text{enc}} = Q\left(\frac{r^3}{R^3}\right)$

So,

$E(4\pi r^2)=\frac{Q r^3}{\varepsilon_0 R^3}$

$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^3}r$

Hence, inside the sphere,

$E \propto r$

So the graph is a straight line through the origin.

3. Field outside the sphere $(r\ge R)$

The whole charge acts as if concentrated at the centre:

$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$

Hence, outside the sphere,

$E \propto \frac{1}{r^2}$

So the graph decreases as an inverse-square curve.

4. At the surface $(r=R)$

From inside:

$E_{\text{in}} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}$

From outside:

$E_{\text{out}} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}$

So the electric field is continuous at $r=R$.

Thus the correct graph must:

• start from $E=0$ at $r=0$,
• increase linearly up to $r=R$,
• then decrease as $1/r^2$ for $r>R$.

5. Matching with options

This corresponds to Option D.

6. Comparison with stored answer

Stored correct answer = D

Derived answer = D

So they agree.