Question

A sphere of radius $R$has total charge $Q$. The volume charge density $\left(\mathrm{C}/{\mathrm{m}}^3\right)$within the sphere is

$\rho ={\rho }_0\left(1-\frac{r}{R}\right)$

This charge density decreases linearly from \(\rho_{0}\) at the center to zero at the edge of the sphere.

a. Show that ${\rho }_0=3Q/\pi R^3$.

b. Show that the electric field inside the sphere points radially outward with magnitude

$E=\frac{Qr}{4\pi {ϵ}_0{R}^3}\left(4-3\frac{r}{R}\right)$

Short answer

a. The constant $\rho 0$of the charge density is found to be $\rho 0=\frac{3Q}{\pi R^3}$

b. The electric field inside the sphere is thus:

$E=\frac{Q}{4\pi \epsilon 0}\frac{1}{R^3}r\left(4-\frac{3r}{R}\right)$

c. The expected boundary condition is satisfied.

Step-by-step solution

part (a) step 1: To find a constant in charge distribution

We use Gauss' Law to find the electric field inside the sphere.

$\Phi =\oint \overrightarrow{E}·\overrightarrow{da}=\frac{Q}{c0}$

Where

$\mathit{\Phi }$is the Electric Flux

$\mathit{E}$is the Electric Field

$d$a is the area

$Q$is the total charge

role="math" localid="1650119288538" $\epsilon 0$is the permittivity

The differential charge d q inside the sphere with charge distribution $\rho$and differential volume $dv$is given by the expression:

$dq=\rho dv$

Since the charge density $\rho$is given by

$\rho =\rho 0\left(1-\frac{r}{R}\right)$

Where

$\mathit{r}$is the radius

$R$is the radius of sphere

The total charge in the sphere can be evaluated with a cylinder as

$dq=\rho 0\left(1-\frac{r}{R}\right)\left(4\pi r^2\right)dr$

Integrating we get:

$Q=\int dq=\int \rho 0\left(1-\frac{r}{R}\right)\left(4\pi r^2\right)dr$

Evaluating the integral between 0 and $R$we get:

$Q=\int dq={\int }_0^R\rho 0\left(1-\frac{r}{R}\right)\left(4\pi r^2\right)dr$

$Q=4\pi R^3\rho 0\left(\frac{1}{12}\right)$

From the above equation we can confirm that

$\rho 0=\frac{3Q}{z{R}^3}$

part (b) step 1: The electric field inside sphere

From part(a) we have $Q=4\pi R^3\rho 0\left(\frac{1}{12}\right)$and $\rho 0=\frac{3Q}{\pi R^3}$

Where

$Q$is the total charge

$r$is the radius

$R$is the radius of sphere

To evaluate the total charge inside the sphere, we have to repeat the process above except that we must replace $R$with $r$.

Thus:

$Q\left(r\right)=\frac{Q}{R^3}{r}^3\left(4-\frac{3r}{R}\right)$

The electric field inside the sphere can be evaluated using Gauss Law with area $4\pi r^2$and charge

$Q\left(r\right)=\frac{Q}{R^3}{r}^3\left(4-\frac{3r}{R}\right):$

$E\left(4\pi r^2\right)=\frac{Q\left(r\right)}{c0}$

Substituting $Q\left(r\right)=\frac{Q}{R^3}{r}^3\left(4-\frac{3r}{R}\right)$we get,

$E\left(4\pi r^2\right)=\frac{Q}{ϵ0}\left(\frac{1}{R^3}{r}^3\left(4-\frac{3r}{R}\right)\right)$

part (c) step 1: To check whether the boundary condition is satisfied

From part(b) we have the electric field inside the sphere

$E=\frac{Q}{4\pi x0}\frac{1}{R^3}r\left(4-\frac{3r}{R}\right)$

Where

$Q$is the total charge

$\mathit{r}$is the radius

$R$is the radius of sphere

$E$is the Electric Field

$\mathit{\epsilon }0$is the permittivity

The above equation at $r=R$evaluates to

$E=\frac{Q}{4\pi z0R^2}$

This is expected. This is expected. The reason this is expected is because it is equal to electric field from a sphere at the surface. This can be evaluated by Gauss law for the total charge density $Q$, spherical area $4\pi R^2$

Substituting all the values in Gauss Law:

$E\left(4\pi R^2\right)=\frac{Q}{{}^{2}0}$

This reduces to $E=\frac{Q}{4xx\left(R^2\right.}$

That is exact equation that is evaluated above for $r=R$. And that is why this is expected.