© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the given potential
The electrostatic potential inside a charged spherical ball is given by
$ \phi = a r^2 + b $
where $r$ is the distance from the center, and $a,b$ are constants.
Step 2: Determine the electric field from the potential
Recall that the electric field $ \mathbf{E} $ is related to the potential $\phi$ by
$ \mathbf{E} = - \nabla \phi $.
In spherical symmetry,
$ E(r) = -\frac{d\phi}{dr}. $
Thus,
\[
E(r) = - \frac{d}{dr}\bigl(a r^2 + b\bigr) = -2 a r.
\]
Step 3: Relate the electric field to the charge density via Gauss's law
Gauss's law in differential form states
\[
\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0},
\]
where $\rho$ is the volume charge density. For a radially symmetric field $E(r)\,\hat{r}$,
\[
\nabla \cdot \mathbf{E} = \frac{1}{r^2} \frac{d}{dr} \bigl(r^2 E(r)\bigr).
\]
Substitute $E(r) = -2 a r$:
\[
\nabla \cdot \mathbf{E}
= \frac{1}{r^2} \frac{d}{dr}\bigl(-2 a r \cdot r^2\bigr)
= \frac{1}{r^2} \frac{d}{dr}\bigl(-2 a r^3\bigr).
\]
Differentiating, we get
\[
\frac{d}{dr} \bigl(-2 a r^3\bigr)
= -6 a r^2.
\]
Hence,
\[
\nabla \cdot \mathbf{E}
= \frac{-6 a r^2}{r^2}
= -6 a.
\]
Step 4: Solve for the charge density
From
$ \nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}, $
we have
\[
-6 a = \frac{\rho}{\varepsilon_0}
\quad \Rightarrow \quad
\rho = -6 a \,\varepsilon_0.
\]
Step 5: State the final answer
Therefore, the charge density inside the spherical ball is
\[
\boxed{-6a\,\varepsilon_0}.
\]
