# What is magnitude of electric field?

## Let $\rho \left(r\right) = \frac{Q \cdot r}{\pi \cdot {R}^{4}}$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q$. for a point ‘p’ inside the sphere at distance $r 1$ from the centre of the sphere, Find the magnitude of electric field. Note : $\rho$ is Volume charge density Please explain elaborately. Thanks in advance :)

Jul 6, 2016

$| \vec{E} | = \frac{Q {\left(r 1\right)}^{2}}{4 \pi {\epsilon}_{0} {R}^{4}}$

#### Explanation:

Ok we're going to use Gauss' law here. Imagine a Gaussian surface at radius r1 inside the sphere. The net flux leaving the surface is only affected by the parts of the sphere inside it.

$\oint \vec{E} \cdot \mathrm{dv} e c \left(A\right) = \frac{{Q}_{e n c}}{{\epsilon}_{0}}$

By symmetry we can see that the electric field vector will always point radially outwards from the centre of the sphere which, incidentally, is the same direction as the area element at all points on our Gaussian surface. We can then write

$\vec{E} \cdot \mathrm{dv} e c \left(A\right) = | \vec{E} | | \mathrm{dv} e c \left(A\right) | \cos \theta$

but $\theta = 0$ so $\vec{E} \cdot \mathrm{dv} e c \left(A\right) = | \vec{E} | | \mathrm{dv} e c \left(A\right) |$

We now have:

$| \vec{E} | \oint | \mathrm{dv} e c \left(A\right) | = \frac{{Q}_{e n c}}{{\epsilon}_{0}}$

ie $| \vec{E} | | \vec{A} | = \frac{{Q}_{e n c}}{{\epsilon}_{0}}$

We now need to work out the charge enclosed by our Gaussian surface. We have a volume charge density, so first we consider a spherical shell of thickness $\mathrm{dr}$. This shell will have charge $\mathrm{dQ}$ given by

$\mathrm{dQ} = \rho 4 \pi {r}^{2} \mathrm{dr}$

$\mathrm{dQ} = \left(a r\right) \cdot 4 \pi {r}^{2} \mathrm{dr}$ where $a = \frac{Q}{\pi {R}^{4}}$

${Q}_{e n c} = 4 a \pi {\int}_{0}^{r 1} {r}^{3} \mathrm{dr}$

${Q}_{e n c} = \frac{Q}{\pi {R}^{4}} \pi {\left(r 1\right)}^{4} = Q {\left(\frac{r 1}{R}\right)}^{4}$

$\implies | \vec{E} | | \vec{A} | = \frac{Q}{{\epsilon}_{0}} {\left(\frac{r 1}{R}\right)}^{4}$

$\implies | \vec{E} | = \frac{Q}{{\epsilon}_{0} \cdot | \vec{A} |} {\left(\frac{r 1}{R}\right)}^{4}$

$| \vec{E} | = \frac{Q}{{\epsilon}_{0} \cdot 4 \pi {\left(r 1\right)}^{2}} {\left(\frac{r 1}{R}\right)}^{4}$

$| \vec{E} | = \frac{Q {\left(r 1\right)}^{2}}{4 \pi {\epsilon}_{0} {R}^{4}}$

I think this makes sense, as at r1 = R it provides the expected electric field outside a sphere. I will admit my EM is a bit rusty though.

Dec 8, 2016

We use Gauss' law here.
Given is volume charge density $\rho = \frac{Q}{\pi {R}^{4}} r$
We observe that by symmetry the electric field vector will always point radially outwards from the center of charged distribution.

Imagine a spherical Gaussian surface of a of radius ${r}_{1}$ inside charged distribution where point $p$ is located. Where ${r}_{1}$ can have values from $0 \text{ to } R$.

We know that net flux leaving the Gaussian surface is contributed by the total charge inside this sphere of radius ${r}_{1}$.
${Q}_{\text{eff}}$ charge enclosed within sphere of radius of ${r}_{1}$ can be found by taking integral from limit $r = 0 \text{ to } {r}_{1}$

Let us we consider charge enclosed in infinitesimal volume $\mathrm{dv}$ having charge $\mathrm{dQ}$ in a spherical shell of thickness $\mathrm{dr}$ at radius $r$
$\mathrm{dQ} = \rho 4 \pi {r}^{2} \mathrm{dr}$
$\implies \mathrm{dQ} = \frac{Q}{\pi {R}^{4}} r \times 4 \pi {r}^{2} \mathrm{dr}$
$\implies \mathrm{dQ} = \frac{Q}{\pi {R}^{4}} r \times 4 \pi {r}^{2} \mathrm{dr}$
$\implies \mathrm{dQ} = \frac{4 Q}{{R}^{4}} {r}^{3} \mathrm{dr}$

Therefore,
$Q \text{eff} = {\int}_{0}^{{r}_{1}} \frac{4 Q}{{R}^{4}} {r}^{3} \mathrm{dr}$
$\implies Q \text{eff} = | \frac{4 Q}{{R}^{4}} {r}^{4} / 4 {|}_{0}^{{r}_{1}}$
$\implies Q \text{eff} = \frac{Q}{{R}^{4}} {r}_{1}^{4}$

As the angle of electric field vector with any element area at the surface of sphere is ${0}^{\circ}$, in the dot product we have $\cos \theta = 1$.

Therefore, from Gauss's law we obtain the electric flux $\Phi$ in this radially symmetrical charged distribution as
Phi=|vecE|cdot"Area of Gaussian Surface"=("Charge "Q_"eff")/epsilon_0

Inserting calculated values we get
$| \vec{E} | \cdot 4 \pi {r}_{1}^{2} = \frac{\frac{Q}{{R}^{4}} {r}_{1}^{4}}{\epsilon} _ 0$
$\implies | \vec{E} | = \left(\frac{Q}{{R}^{4}} {r}_{1}^{4}\right) \frac{1}{4 \pi {r}_{1}^{2}} \frac{1}{\epsilon} _ 0$
$\implies | \vec{E} | = \frac{Q {r}_{1}^{2}}{4 \pi {\epsilon}_{0} {R}^{4}}$