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

 

 

 

 

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