Question

A solid ball of radius $r_b$ has a uniform charge density .

A)

What is the magnitude of the electric field at a distance $r>r_b$ from the center of the ball?

Express your answer in terms of, $r_b$ , ,and $epsilon_0$ .

=

Part B

What is the magnitude of the electric field at a distance $r<r_b$ from the center of the ball?

Express your answer in terms of, , $r_b$ , and $epsilon_0$ .

=

Part C

Let represent the electric field due to the charged ballthroughout all of space. Which of the following statements aboutthe electric field are true?

1. $E(0)=0$ .
2. $E(r_{\rm b})=$ .
3. $\lim_{r\to\infty}E(r)=0$ .
4. The maximum electric field occurs when $r=0$ .
5. The maximum electric field occurs when $r=r_{\rm b}$ .
6. The maximum electric field occurs as $r\to\infty.$

Answer 1

Concepts and reason

The concept related to solve this problem is the gauss’s law. First, calculate the magnitude of the electric field at a distance $r > {r_b}$ from the center of the ball. Later, calculate the magnitude of the electric field at a distance $r < {r_b}$ from the center of the ball. Finally, explain the electric field due to the charged ball throughout all of space and identify which of statement about the electric field is true.

Fundamentals

Gauss’s Law: The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by${\varepsilon _0}$. Mathematically, the statement can be written as,

$\oint {\vec E \cdot d\vec A} = \frac{q}{{{\varepsilon _0}}}$

Here, $\vec E$ is the electric field, $d\vec A$ is the infinitesimal surface element, q is the charge enclosed in the sphere, and ${\varepsilon _0}$ is the permittivity of vacuum.

The expression for the electric field is,

$\begin{array}{c}\\E\left( {4\pi {r^2}} \right) = \frac{q}{{{\varepsilon _0}}}\\\\E = \frac{1}{{4\pi {r^2}}}\frac{q}{{{\varepsilon _0}}}\\\end{array}$

Here, r is the radius of the sphere.

The charge enclosed in the sphere is written as:

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

Here, $\rho$ is the charge density of the sphere.

Part-A

Calculate the magnitude of the electric field at a distance from the center of the ball.

The expression for the gauss law is,

$\begin{array}{l}\\\int {E \cdot dA} = \frac{q}{{{\varepsilon _0}}}\\\\E\int {dA = \frac{q}{{{\varepsilon _0}}}} \\\end{array}$

The area of the sphere is $4\pi {r^2}$.

Substitute $4\pi {r^2}$for $\int {dA}$ in the above equation.

$\begin{array}{c}\\E\left( {4\pi {r^2}} \right) = \frac{q}{{{\varepsilon _0}}}\\\\E = \frac{1}{{4\pi {r^2}}}\frac{q}{{{\varepsilon _0}}}\\\end{array}$

Here, q is the charge enclosed in the sphere, r is the radius of the sphere, and ${\varepsilon _0}$ is the permittivity of free space.

The volume density of the sphere is written as:

$\begin{array}{l}\\\rho = \frac{q}{V}\\\\q = \rho V\\\end{array}$

The volume of the sphere is $\frac{4}{3}\pi {r^3}$.

Rewrite the above equation as follows:

$q = \rho \left( {\frac{4}{3}\pi {r_b}^3} \right)$

Substitute q value in the gauss law equation. The equation of electric field is written as:

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

Part-B

Calculate the charge enclosed in the sphere by using gauss law.

The charge enclosed in the sphere is written as:

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

Now, the equation for the electric field is written as:

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

Part-C

From the above two equations, when the radius r is zero then the electric field will be zero. At infinite distance, the electric field is zero. Maximum electric field is produced at a distance of ${r_b}$. Hence, the correct options are 1, 3, 5.

Ans: Part-A

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

Part-B

The magnitude of the electric field at a distance $r < {r_b}$ from the center of the ball is $E = \frac{{\rho r}}{{3{\varepsilon _0}}}$

Part-C

The following statements are true. $E\left( 0 \right) = 0$, $\mathop {\lim }\limits_{r \to \infty } E\left( r \right) = 0$, and the maximum electric field occurs when $r = {r_b}$.