Constant Electric Field

Consider a sphere with radius $R$ centered on the origin of the coordinate system (i.e., the point where $r=0$ ) and let that be our Gaussian surface. The electric flux at a point on the surface depends only on the enclosed electric charge and on the permissivity of the medium. Therefore, the charges at a distance greater than $R$ from the origin can be ignored when evaluating the electric field at a distance $R$ from the origin (i.e., the electric field on the sphere of radius $R$ ).

The charge (consider it positive) enclosed by the sphere is $q = \int\limits_0^R \rho (r)dV$ , where $dV$ is a volume element (i.e., the volume of the infinitesimal shell between the spheres of radius $r$ and $r+dr$ ). It follows that $dV = \frac {4\pi}{3}(r+dr)^3 -\frac {4\pi}{3}r^3$ . Neglecting the terms in ( $dr)^2$ and $(dr)^3$ , the result is $dV = 4\pi r^2dr$ . Now it's possible to find the value of the enclosed charge:

$q = \int\limits_0^R Cr^{\alpha}4\pi r^2dr = 4\pi C\int\limits_0^Rr^{2+\alpha}dr = \left. 4\pi C \frac {r^{3+\alpha}}{3+\alpha} \right|_0^R = 4\pi C \frac {R^{3+\alpha}}{3+\alpha}$ (this is an alternative to computing a triple integral)

Gauss's Law states that:

$\oint \overrightarrow{E}\cdot d\overrightarrow{S} = \frac{4\pi C}{\epsilon_{0}} \frac {R^{3+\alpha}}{(3+\alpha)}$

Because of the spherical symmetry, the electric field is constant on all points of the Gaussian surface. Also, since $\overrightarrow{E}$ and $d\overrightarrow{S}$ are parallel on all points of the sphere (both are radially outward), $\oint \overrightarrow{E}\cdot d\overrightarrow{S} = ES = E4\pi R^2$ , which implies:

$E = \frac{C}{\epsilon_{0}} \frac {R^{1+\alpha}}{(3+\alpha)}$

If the electric field is constant in magnitude, its value must be the same at every point in space, all of which correspond to a value of $R$ . Since $C$ and $\epsilon_{0}$ are constant, the expression of the electric field cannot have a term in $R$ . In other words, the exponent of $R$ must be zero. That leads to $1+\alpha=0 \rightarrow \alpha= -1$ and $E = \frac{C}{2\epsilon_{0}}$

${ Q }_{ Tot }=\rho *v=(C{ r }^{ \alpha })\frac { 4 }{ 3 } \pi { r }^{ 3 }$

Since we want the electric field to be constant We take it out of the integral

$E\oint { dA= } \frac { C{ r }^{ \alpha } }{ e } \frac { 4 }{ 3 } \pi { r }^{ 3 }\\ E=\frac { 1 }{ 3 } \frac { C{ r }^{ \alpha } }{ e } \pi { r }\\ E=\frac { 1 }{ 3e } C\pi ({ r }^{ \alpha +1 })\quad \Longrightarrow \alpha =-1$

E=Q/Eo dQ=spheric surface * dR * density of charge(depending of radius) so I want the electric field to be constant in the space, so dE/dR = 0 Solving this you get the answer.

electric field is directly propotional to current density* radius(By dimensional analysis) so C.D must be proptional to 1/r to make E constant

By Gauss' Law, $\oint_S E \cdot dA = \frac{q_{enc}}{\varepsilon _0 }$ .

In the case of a spherically symmetric distribution, we have $E(4\pi r^2)=\frac{\frac{4}{3}\pi r^3 \rho(r)}{\varepsilon_0}$ , which yields $E=\frac{\frac{1}{3} r (Cr^{\alpha}) }{\varepsilon_0}$ .

In order for $E$ to be constant in magnitude, $\alpha = \fbox{-1}$ .

Let us consider a spherical Gaussian surface of radius R. Applying Gauss Law yields $\Phi=4\pi R^{2} E(R)= \frac{Q(R)}{\epsilon_{0}}$ where $Q(R)$ is the charge enclosed by surface. This charge can be computed as follows: $Q(R)=\int \rho(r) dV = 4\pi \int_{0}^{R} r^{2} \rho(r) dr= 4\pi C \frac{R^{\alpha+3}}{\alpha+3}.$ Comparing the above equation we find that for $\alpha=-1$ the electric field is constant and it is given by $E(r)=\frac{C}{2\epsilon_{0}}.$ The conclusion is that a charge density of the form $\rho(r) \sim \frac{1}{r}$ generates a spherically symmetic field of constant magnitude.

Gauss' law in differential form is $\vec\nabla \cdot\vec E = \frac \rho {\varepsilon_0}.$ Because of symmetry, $\vec E$ is radial: $\vec E = E_r\:\hat r$ . Using spherical coordinates, $\frac 1{r^2}\ \frac{d}{dr}(r^2E_r) = \frac \rho{\varepsilon_0}.$ If $E_r$ is constant, this reduces to $\frac {2E_r r}{r^2} = \frac \rho{\varepsilon_0}\ \ \ \therefore\ \ \ \rho = \frac{2\varepsilon_0 E_r}r,$ from which it is clear that $C = 2\varepsilon_0 E_r$ and $\alpha = \boxed{-1}$ .