Question

What is the net electric flux through the cylinder of FIGURE?

Short answer

The net electric flux through the cylinder is ${\mathrm{\Phi }}_{\mathrm{e}}=113\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$

Step-by-step solution

Introduction

The number of electric lines of force (or equipotential lines) that cross a given region is the characteristic of an electric field named electric flux. Electric field lines are thought to start with positive electric charges and end with negative ones.

Explanation

The electricity field that travels through a closed surface is called to as the electric flux. The electric flux through a surface is proportional to the charge inside the surface, according to Gauss's law, which is given by equation $(24.18)$in the form

Equation $1$

role="math" localid="1649250954712" ${\mathrm{\Phi }}_{\mathrm{e}}=\text{∮}\overrightarrow{E}\text{X}d\overrightarrow{A}=\frac{Q_{\text{in}}}{{ϵ}_o}$

The electric flow rate is determined by the charge inside the closed surface, as indicated. Any flux related to charges outside the closed surface is zero, hence to compute the flux, we use the negative charge inside the cylinder. Outside the Gaussian surface are $-1nC$, $+100nC$, and $-100nC$. The inert charge is a charge that has no impact on the body.

$$\begin{gathered} Q_{\text{in}}=(+1\mathrm{nC})\left(\frac{1\times {10}^{-9}\mathrm{C}}{\mathrm{nC}}\right) \\ =+1\times {10}^{-9}\mathrm{C} \end{gathered}$$

Finally, to get ${\mathrm{\Phi }}_{\mathrm{e}}$, we plug the values for $Q_{\text{in}}\text{and}{ϵ}_O$into equation ($1$).

${\mathrm{\Phi }}_{\mathrm{e}}=\frac{Q_{\mathrm{in}}}{{ϵ}_o}$

$=\frac{1\times {10}^{-9}\mathrm{C}}{8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}\cdot {\mathrm{m}}^2}$

$=113\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$