Solving for the electric field using Gauss’ theorem | Quantum Science Philippines

Solving for the electric field using Gauss’ theorem

Bianca Rae B. Sambo

Problem 1.4 (Classical Electrodynamics, 3rd Edition by Jackson)

Each of the three charged spheres of radius a has a total charge Q. One is conducting, one has a uniform charge density within its volume and one having a spherically symmetric charge density that varies radially as $r^n$ where (r>-3). Use Gauss’ theorem to obtain the electric fields both inside and outside the sphere.

SOLUTION:

A. Conducting Sphere

A.1 INSIDE

Note that no charge resides inside a conducting sphere. All charges reside in the outer surface thus $\oint{E}\bullet{nda}=0$ which implies that $\vec{E}=0$   for r<a

A.2 OUTSIDE

Let r be the distance from the center of sphere a. $\oint{E}\bullet{nda}=\frac{Q}{\epsilon_0}$ $\oint{Edacos{\theta}}=\frac{Q}{\epsilon_0}$ $\oint{Er^2d\Omega}=\frac{Q}{\epsilon_0}$ where $d\Omega$ is the solid angle. ${E}{4\pi}r^2 =\frac{Q}{\epsilon_0}$ $\vec{E}=\frac{Q}{{4\pi}{\epsilon_0}r^2}\hat{r}$          where ${r}\geq{a}$

B. SPHERE WITH UNIFORM CHARGE DENSITY

Since Q is specified to be the total charge then we can get an expression for $\rho$ $\rho=\frac{Q}{volume}=\frac{3Q}{{4\pi}a^3}$ so that

B.1 INSIDE SPHERE B

Let b be the distance from the center of the sphere. $\oint{E}\bullet{nda}=\frac{1}{\epsilon_0}{\int_v{\rho dV}}=\frac{3Q}{{4\pi}{\epsilon_0}a^3}\int{b^2db}\int{sin\theta{d\theta}}\int{d\phi}$

b is evaluated from 0 to r where r<a, $\theta$ from 0 to $\pi$ and $\phi$ from o to $2\pi$ $\oint{E{r^2}{d\Omega}}=\frac{3Q}{{4\pi}{\epsilon_0}a^3} {4\pi} \frac{r^3}{3}$ $E{4\pi}{r^2}=\frac{3Q}{{4\pi}{\epsilon_0}a^3} {4\pi} \frac{r^3}{3}$ $\vec{E} =\frac{Q}{{4\pi}{\epsilon_0}a^3} r\hat{r}$   but $\vec{r}=r\hat{r}$ $\vec{E}=\frac{Q}{{4\pi}{\epsilon_0}a^3}\vec{r}$

B.2 OUTSIDE THE SPHERE

Outside the sphere, the total charge enclosed is still Q. $q_{enc}=\int_v{{\rho}dV}=\frac{3Q}{{4\pi}a^3} {4\pi} \frac{r^3}{3}$

where r is evaluated from o to a $q_{enc}=Q$

so that $\oint{E{\bullet}nda}=\oint{E{cos\theta}da}=\oint{E{r^2}{d\Omega}}=\frac{Q}{\epsilon_0}$ $\vec{E}=\frac{Q}{{4\pi}{\epsilon_0}r^2}\hat{r}$          where ${r}\geq{a}$

C. SPHERE WITH ${\rho}{\propto}{r^n}$

The charge density has the form ${\rho}=A{r^n}$ where A is a constant. Now the total charge for sphere C should be Q thus $Q=\int{{\rho}dV}=A\int{{r^n}{r^2}{dr}}\int{{sin\theta}d\theta}\int{{d\Omega}}$ $Q=A\int{{r^{n+2}}{dr}}\int{{sin\theta}d\theta}\int{{d\Omega}}=A\frac{r^{n+3}}{(n+3)}{4\pi}=A{\frac{{4\pi}}{(n+3)}}{a^{n+3}}$

since r is evaluated from o to a

so that $A=\frac{(n+3)Q}{{4\pi}a^{n+3}}$

and the charge density can be written as $\rho=\frac{(n+3)Q}{{4\pi}a^{n+3}} r^n$ where r is the distance from the center of sphere C

C.1 INSIDE SPHERE C $\oint{E{\bullet}nda}=\frac{1}{\epsilon_0}\int_v{\frac{(n+3)Q}{{4\pi}a^{n+3}} r^n}dV$ $\oint{E{cos\theta}da}=\frac{1}{\epsilon_0}{4\pi}\frac{(n+3)Q}{{4\pi}a^{n+3}}\int{r^{n+2}}dr$ $E{4\pi}{r^2}=\frac{Q{r^{n+3}}}{{\epsilon_0}{a^{n+3}}}$ $\vec{E}=\frac{Q{r^{n+1}}}{{4\pi}{\epsilon_0}{a^{n+3}}}\hat{r}$  where r<a

C.2 OUTSIDE THE SPHERE $\oint{E{\bullet}nda}=\frac{1}{\epsilon_0}\int_v{\frac{(n+3)Q}{{4\pi}a^{n+3}} r^n}dV$ $\oint{E{cos\theta}da}=\frac{1}{\epsilon_0}{4\pi}\frac{(n+3)Q}{{4\pi}a^{n+3}}\int{r^{n+2}}dr$ where r is evaluated from o to a $E{4\pi}{r^2}=\frac{Q{a^{n+3}}}{{\epsilon_0}{a^{n+3}}}$ $\vec{E}=\frac{Q}{{4\pi}{\epsilon_0}{r^2}}\hat{r}$ where ${r}\geq{a}$