StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q. 61 - Excercises And Problems

Expert-verifiedFound in: Page 725

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

A spherical ball of charge has radius R and total charge Q. The electric field strength inside the ball $(r\le R)isE\left(r\right)={r}^{4}{E}_{\mathrm{max}}/{R}^{4}.\phantom{\rule{0ex}{0ex}}$

a. What is${E}_{\mathrm{max}}$ in terms of Q and R ?

b. Find an expression for the volume charge density \rho(r) inside the ball as a function of r.

c. Verify that your charge density gives the total charge Q when integrated over the volume of the ball.

$\rho =\frac{6Q}{4\pi {R}^{3}}\left(\frac{{r}^{3}}{{R}^{3}}\right)$a.The maximum electric Field is

$E\mathrm{max}=\frac{Q}{4\pi \epsilon 0{R}^{2}}$

b. So the charge density is $\rho =\frac{6Q}{4\pi {R}^{3}}\left(\frac{{r}^{3}}{{R}^{3}}\right)$

c. We can see that the charge density does equal Q when integrated over the volume of the sphere, given that's how it was derived.

The maximum electric field is

$Emax=\frac{Q}{4\pi \epsilon 0{R}^{2}}$

Where

Emax is the maximum Electric Field

R is the radius of sphere

Q is the total charge

e0 is the permittivity

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

$\Phi =\oint \overrightarrow{E}\xb7\overrightarrow{da}=\frac{Q}{\epsilon 0}$

Where

$\Phi $ is the Electric Flux

E is the Electric Field

d a is the area

Q is the total charge

$\epsilon $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 $4\pi {R}^{2}$

" Substituting all the values in Gauss Law:

$Emax\left(4\pi {R}^{2}\right)=\frac{Q}{t0}$

This reduces to

${E}_{\mathrm{max}}=\frac{Q}{4\pi \xb70{R}^{2}}$

The charge density is

$\rho =\frac{6Q}{4\pi {R}^{3}}\left(\frac{{r}^{3}}{{R}^{3}}\right)$

Where

r is the radius

R is the radius of sphere

Q is the total charge

$\epsilon 0$ is the permittivity

From part(a) the maximum electric field is

$Emax=\frac{Q}{4\pi 20{R}^{2}}$

Where

Emax is the maximum Electric Field

R is the radius of sphere

Q is the total charge

$\epsilon 0$ is the permittivity

Let us assume a charge density

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

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:

$dq=\rho dv$

The total charge in the sphere can be evaluated as

$dq=\rho 0\left(\frac{{r}^{3}}{{R}^{3}}\right)\left(4\pi {r}^{2}\right)dr$

Integrating we get:

$Q=\int dq=\rho 0\left(\frac{{r}^{3}}{{R}^{3}}\right)\left(4\pi {r}^{2}\right)dr$

Evaluating the integral between 0 and R we get:

$Q=\frac{4\pi p0{R}^{6}}{6{R}^{3}}$

From the above equation we can confirm that

$\rho 0=\frac{6Q}{4\pi {R}^{3}}$

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

Thus:

$Q\left(r\right)=\frac{4\pi p0{r}^{6}}{6{R}^{3}}$

The electric field inside the sphere can be evaluated as

$E=\frac{Q\left(r\right)}{4\pi {r}^{2}\xb70{R}^{2}}\frac{{r}^{6}}{{R}^{4}}$

Which is nothing but

$E=Emax\frac{{r}^{2}}{{k}^{3}}$

Which is nothing but

$E=E\mathrm{max}\frac{{r}^{4}}{{R}^{4}}$

The total charge is obtained upon integrating

Explanation of Solution

From part(b) the charge density is

$\rho =\frac{6Q}{4\pi {R}^{3}}\left(\frac{{r}^{3}}{{R}^{3}}\right)$

Where

r is the radius

R is the radius of sphere

Q is the total charge

$\epsilon 0$is the permittivity

$Q=\int dq=\int \rho d\nu $

Where

d v is the differential volume

Q is the total charge

$\rho $ is the charge density

From the above equation, substitute $\rho =\frac{6Q}{4\pi {R}^{3}}\left(\frac{{r}^{3}}{{R}^{3}}\right)$, and the differential volume $dv=\left(4\pi {r}^{2}\right)dr$and limits 0 and R. The total charge is:

$Q={\int}_{0}^{R}\rho dv={\int}_{0}^{R}\frac{6Q}{4\pi {R}^{3}}\left(\frac{{r}^{3}}{{R}^{3}}\right)\left(4\pi {r}^{2}\right)d$

$Q={\int}_{0}^{R}\frac{6Q}{4\pi {R}^{3}}\left(\frac{{r}^{3}}{{R}^{3}}\right)\left(4\pi {r}^{2}\right)dr={\int}_{0}^{R}\frac{6Q{r}^{5}}{{R}^{6}}dr=\frac{6Q{R}^{6}}{6{R}^{6}}=Q$

94% of StudySmarter users get better grades.

Sign up for free