Question

The electric field is horizontal and has the values indicated on the surface of cylinder as shown in Figure 21.65. What can you deduce from this pattern of electric field? Include a numerical result.

Short answer

We can deduce from the above electric field distribution that cylinder contains the charge $-5.31\times {10}^{-9}\text{C}$ between surfaces of cylinder.

Step-by-step solution

Identification of given data

The length of cylinder is$l=15cm$

The radius of cylinder is $r=5cm$.

The magnitude of incoming electric field on the left side of cylinder is $E_1=1600N/C$.

The magnitude of incoming electric field on the left side of cylinder is $E_2=1000N/C$.

Conceptual Explanation

The Gauss law is applied for the cylinder to find the net charge inside the cylinder for incoming and outgoing electric field of cylinder. For this net flux for cylinder is equated to the charge inside surface divided by permittivity of free space.

Determination of charge enclosed inside the cylinder

The surface area of the cylinder is given as:

$A=2\pi rl$

Substitute all the values in the above equation.

$$\begin{gathered} A=2\pi \left[\left(5\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right]\left[\left(15\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right] \\ A=0.0471{\text{m}}^2 \end{gathered}$$

Apply the Gauss’s law to find the amount of charge inside cylinder:

$\left(E_2-E_1\right)A=\frac{q}{{\epsilon }_0}$

Here, ${\epsilon }_0$ is the permittivity of box and its value is$8.854\times {10}^{-12}{\text{C}}^2/\text{N}·{\text{m}}^2$ .

Substitute all the values in the above equation.

$\begin{array}{rcl}\left(1000\text{N}/\text{C}-1600\text{N}/\text{C}\right)& =& \frac{q}{8.854\times {10}^{-12}{\text{C}}^2/\text{N}·{\text{m}}^2}\\ q& =& -5.31\times {10}^{-9}\text{C}\\ & & \end{array}$

Therefore, the amount of charge inside the cylinder is $-5.31\times {10}^{-9}\text{C}$.