91Ó°ÊÓ

We use Gauss' Law to find the electric field inside the slab of a given thickness.

$\mathrm{Φ}=âˆ\circledR \stackrel{→}{E}·\stackrel{→}{da}=\frac{Q}{\mathrm{Î\mu }0}$

Where

$\mathrm{Φ}$is the Electric Flux

E is the Electric Field

d a is the area

Q is the total charge

$\mathrm{Î\mu }$0 is the permittivity

The Emax or the maximum electric field is equal to the electric field at the surface of the sphere of radius R. It can be evaluated by Gauss law for the total charge density Q and spherical area$4\mathrm{Ï€}{R}^2$

Substituting all the values in Gauss Law:

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

This reduces to

$Emax=\frac{Q}{4\mathrm{Ï€}\mathrm{Î\mu }0R^2}$

From part(a) the maximum electric field is

$Emax=\frac{Q}{4\mathrm{Ï€}\mathrm{Î\mu }0R^2}$

Emax is the maximum Electric Field

R is the radius of sphere

Q is the total charge

$\mathrm{Î\mu }$0 is the permittivity

Let us assume a charge density

$\mathrm{ÒÏ}=\mathrm{ÒÏ}0\left(\frac{r^3}{R^3}\right)$

Where

r is the radius

R is the radius of sphere

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

$dq=\mathrm{ÒÏ}dv$

The total charge in the sphere can be evaluated as

$dq=\mathrm{ÒÏ}0\left(\frac{r^3}{R^3}\right)\left(4\mathrm{Ï€}{r}^2\right)dr$

Integrating we get:

$Q=∫dq=\mathrm{ÒÏ}0\left(\frac{r^3}{R^3}\right)\left(4\mathrm{Ï€}{r}^2\right)dr$

Evaluating the integral between 0 and $\mathrm{R}$we get:

$Q=\frac{4\mathrm{Ï€}\mathrm{ÒÏ}0R^6}{6R^3}$

From the above equation we can confirm that

$\mathrm{ÒÏ}0=\frac{6Q}{4\mathrm{Ï€}{R}^3}$

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

$Q\left(r\right)=\frac{4\mathrm{Ï€}\mathrm{ÒÏ}0r^6}{6R^3}$

The electric field inside the sphere can be evaluated as

$E=\frac{Q\left(r\right)}{4\mathrm{Ï€}{r}^2\mathrm{Î\mu }0R^2}\frac{r^6}{R^4}$

Which is nothing but

$E=Emax\frac{r^4}{R^4}$

From part(b) the charge density is

$\mathrm{ÒÏ}=\frac{6Q}{4\mathrm{Ï€}{R}^3}\left(\frac{r^3}{R^3}\right)$

Where

r is the radius

R is the radius of sphere

Q is the total charge

$\mathrm{Î\mu }$0 is the permittivity

The total charge is given by

$Q=∫dq=∫\mathrm{ÒÏ}dv$

Where

d v is the differential volume

Q is the total charge

$\mathrm{ÒÏ}$is the charge density

From the above equation, substitute

$\mathrm{ÒÏ}=\frac{6Q}{4\mathrm{Ï€}{R}^3}\left(\frac{r^3}{R^3}\right)$

and the differential volume $dv=\left(4\mathrm{Ï€}{r}^2\right)dr$and limits 0 and R. The total charge is:

$Q={∫}_0^R\mathrm{ÒÏ}dv={∫}_0^R\frac{6Q}{4\mathrm{Ï€}{R}^3}\left(\frac{r^3}{R^3}\right)\left(4\mathrm{Ï€}{r}^2\right)dr$

$Q={∫}_0^R\frac{6Q}{4\mathrm{π}{R}^3}\left(\frac{r^3}{R^3}\right)\left(4\mathrm{π}{r}^2\right)dr={∫}_0^R\frac{6Q{r}^5}{R^6}dr=\frac{6Q{R}^6}{6R^6}=Q$