Question

A solid ball of radius rb has a uniform charge density ρ.

Part A

What is the magnitude of the electric field E(r) at a distance r>rb from the center of the ball?

Express your answer in terms of ρ, rb, r, and ϵ0.

Part B

What is the magnitude of the electric field E(r) at a distance r<rb from the center of the ball?

Express your answer in terms of ρ, r, rb, and ϵ0.

Part C

Let E(r) represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true?

Check all that apply.

Check all that apply.

	E(0)=0.
	E(rb)=0.
	limr→∞E(r)=0.
	The maximum electric field occurs when r=0.
	The maximum electric field occurs when r=rb.
	The maximum electric field occurs as r→∞.

Answer 1

Concepts and reason

The concept required to solve the given problem is of Gauss’s law.

Initially, for part A, apply the Gauss’s law to find the magnitude of the electric field E(r) at a distance r>r_b from the center of the ball. Here, r_b is the radius of the solid ball. Then for part B, apply the Gauss’s law to find

the electric field E(r) at a distance r<r_b from the center of the ball. Then for part C, find out the true statements about the electric field.

Fundamentals

According to the Gauss's law, the net flux of an electric field in a closed surface is directly proportional to the charge enclosed. It is mathematically given by,

$\oint {\vec E.d\vec A} = \frac{Q}{{{\varepsilon _0}}}$

So,

$E = \frac{Q}{{{\varepsilon _0}A}}$

Here, E is the electric field, A is the area, Q is the charge, and ${\varepsilon _0}$ is the permittivity of free space.

The relation of charge and charge density is given by,

$Q = \rho V$

Here, $\rho$ is the charge density, Q is the charge, and V is the volume.

(A)

The Gauss’s law is given by,

$E\left( r \right) = \frac{Q}{{{\varepsilon _0}A}}$

But, $Q = \rho V$ . Thus, substitute $\rho V$ for Q in the above equation.

$E\left( r \right) = \frac{{\rho V}}{{{\varepsilon _0}A}}$

Here, substitute $\frac{4}{3}\pi r_b^3$ for the volume of sphere in the above equation.

$E\left( r \right) = \frac{{\rho \left( {\frac{4}{3}\pi r_b^3} \right)}}{{{\varepsilon _0}A}}$

Now, substitute $4\pi {r^2}$ for A in the above equation.

$\begin{array}{c}\\E\left( r \right) = \frac{{\rho \left( {\frac{4}{3}\pi r_b^3} \right)}}{{{\varepsilon _0}\left( {4\pi {r^2}} \right)}}\\\\ = \frac{{\rho r_b^3}}{{3{\varepsilon _0}{r^2}}}\\\end{array}$

The magnitude of the electric field E(r) at a distance r>r_b from the center of the ball is $E\left( r \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}{r^2}}}$ .

(B)

The Gauss’s law is given by,

$E\left( r \right) = \frac{Q}{{{\varepsilon _0}A}}$

But, $Q = \rho V$ . Thus, substitute $\rho V$ for Q in the above equation.

$E\left( r \right) = \frac{{\rho V}}{{{\varepsilon _0}A}}$

Here, substitute $\frac{4}{3}\pi {r^3}$ for the volume of sphere in the above equation.

$E\left( r \right) = \frac{{\rho \left( {\frac{4}{3}\pi {r^3}} \right)}}{{{\varepsilon _0}A}}$

Now, substitute $4\pi {r^2}$ for A in the above equation.

$\begin{array}{c}\\E\left( r \right) = \frac{{\rho \left( {\frac{4}{3}\pi {r^3}} \right)}}{{{\varepsilon _0}\left( {4\pi {r^2}} \right)}}\\\\ = \frac{{\rho r}}{{3{\varepsilon _0}}}\\\end{array}$

The magnitude of the electric field E(r) at a distance rb from the center of the ball is $E\left( r \right) = \frac{{\rho r}}{{3{\varepsilon _0}}}$ .

(C.1)

The equation used here is,

$E\left( r \right) = \frac{{\rho \left( {\frac{4}{3}\pi {r^3}} \right)}}{{{\varepsilon _0}\left( {4\pi {r^2}} \right)}}$

Substitute 0 for r in equation $E\left( r \right) = \frac{{\rho \left( {\frac{4}{3}\pi {r^3}} \right)}}{{{\varepsilon _0}\left( {4\pi {r^2}} \right)}}$ as follows:”

$\begin{array}{c}\\E\left( 0 \right) = \frac{{\rho \left( {\frac{4}{3}\pi {{\left( 0 \right)}^3}} \right)}}{{{\varepsilon _0}\left( {4\pi {{\left( 0 \right)}^2}} \right)}}\\\\ = 0\\\end{array}$

So, $E\left( 0 \right) = 0$ is true.

(C.2)

The equation used here is,

$E\left( r \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}{r^2}}}$

Substitute r_b for r in equation $E\left( r \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}{r^2}}}$ as follows:

$\begin{array}{c}\\E\left( {{r_b}} \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}r_b^2}}\\\\ = \frac{{\rho {r_b}}}{{3{\varepsilon _0}}}\\\end{array}$

So, $E\left( {{r_b}} \right) \ne 0$ .

Thus, $E\left( {{r_b}} \right) = 0$ is false.

(C.3)

The equation used here is,

$E\left( r \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}{r^2}}}$

Substitute infinity for r in equation $E\left( r \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}{r^2}}}$ as follows:

$\begin{array}{c}\\E\left( \infty \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}r}}\\\\ = \frac{{\rho {r_b}}}{{3{\varepsilon _0}\left( \infty \right)}}\\\\ = 0\\\end{array}$

Thus, $E\left( \infty \right) = 0$ is true.

(C.4)

The equation used here is,

$E\left( r \right) = \frac{{\rho \left( {\frac{4}{3}\pi {r^3}} \right)}}{{{\varepsilon _0}\left( {4\pi {r^2}} \right)}}$

Substitute 0 for r in equation $E\left( r \right) = \frac{{\rho \left( {\frac{4}{3}\pi {r^3}} \right)}}{{{\varepsilon _0}\left( {4\pi {r^2}} \right)}}$ as follows:

$\begin{array}{c}\\E\left( 0 \right) = \frac{{\rho \left( {\frac{4}{3}\pi {{\left( 0 \right)}^3}} \right)}}{{{\varepsilon _0}\left( {4\pi {{\left( 0 \right)}^2}} \right)}}\\\\ = 0\\\end{array}$

So, $E\left( 0 \right) = 0$ .

Thus, the maximum electric field occurs when r=0 is false.

(C.5)

The equation used here is,

$E\left( r \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}{r^2}}}$

Substitute r_b for r in equation $E\left( r \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}{r^2}}}$ as follows:

$\begin{array}{c}\\E\left( {{r_b}} \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}r_b^2}}\\\\ = \frac{{\rho {r_b}}}{{3{\varepsilon _0}}}\\\end{array}$

So, $E\left( {{r_b}} \right) \ne 0$ .

Thus, the maximum electric field occurs when r=r_b is true.

(C.6)

The equation used here is,

$E\left( r \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}{r^2}}}$

Substitute infinity for r in equation $E\left( r \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}{r^2}}}$ as follows:

$\begin{array}{c}\\E\left( \infty \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}r}}\\\\ = \frac{{\rho {r_b}}}{{3{\varepsilon _0}\left( \infty \right)}}\\\\ = 0\\\end{array}$

Thus, the maximum electric field occurs as r→∞ is false.

Ans: Part A

The magnitude of the electric field E(r) at a distance r>r_b from the center of the ball is $E\left( r \right) = \frac{{\rho r_b^3}}{{3{\varepsilon _0}{r^2}}}$ .

Part B

The magnitude of the electric field E(r) at a distance rb from the center of the ball is $E\left( r \right) = \frac{{\rho r}}{{3{\varepsilon _0}}}$ .

Part C.1

The given statement is true.

Part C.2

The given statement is false.

Part C.3

The given statement is true.

Part C.4

The given statement is false.

Part C.5

The given statement is true.

Part C.6

The given statement is false.