91Ó°ÊÓ

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

$\mathrm{Φ}=âˆ\circledR \stackrel{→}{E}·\stackrel{→}{da}=\frac{Q}{c0}$

Where

$\mathit{Φ}$is the Electric Flux

$\mathit{E}$is the Electric Field

$d$a is the area

$Q$is the total charge

role="math" localid="1650119288538" $\mathrm{Î\mu }0$is the permittivity

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

$dq=\mathrm{ÒÏ}dv$

Since the charge density $\mathrm{ÒÏ}$is given by

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

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

$dq=\mathrm{ÒÏ}0\left(1-\frac{r}{R}\right)\left(4\mathrm{Ï€}{r}^2\right)dr$

Integrating we get:

$Q=∫dq=∫\mathrm{ÒÏ}0\left(1-\frac{r}{R}\right)\left(4\mathrm{Ï€}{r}^2\right)dr$

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

$Q=∫dq={∫}_0^R\mathrm{ÒÏ}0\left(1-\frac{r}{R}\right)\left(4\mathrm{Ï€}{r}^2\right)dr$

$Q=4\mathrm{Ï€}{R}^3\mathrm{ÒÏ}0\left(\frac{1}{12}\right)$

From the above equation we can confirm that

$\mathrm{ÒÏ}0=\frac{3Q}{z{R}^3}$

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

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 $4\mathrm{Ï€}{r}^2$and charge

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

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

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

$E\left(4\mathrm{Ï€}{r}^2\right)=\frac{Q}{\mathrm{Ï\mu }0}\left(\frac{1}{R^3}{r}^3\left(4-\frac{3r}{R}\right)\right)$

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

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

Where

$Q$is the total charge

$\mathit{r}$is the radius

$R$is the radius of sphere

$E$is the Electric Field

$\mathit{Î\mu }0$is the permittivity

The above equation at $r=R$evaluates to

$E=\frac{Q}{4\mathrm{Ï€}z0R^2}$

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 $4\mathrm{Ï€}{R}^2$

Substituting all the values in Gauss Law:

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

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.