Question

$55.3$million excess electrons are inside a closed surface. What is the net electric flux through the surface?

Short answer

${\mathrm{\Phi }}_{\mathrm{e}}=-1\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$is the net electric flux which is through the surface.

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 $1$

localid="1649251670705" ${\mathrm{\Phi }}_{\mathrm{e}}=\text{∮}\overrightarrow{E}\text{X}d\overrightarrow{A}=\frac{Q_{\text{in}}}{{ϵ}_o}$

The electric flow is determined by the charge inside the closed surface, as indicated. There is no flow owing to charges outside the closed surface. The amount of ions is revealed to us by the $n=55.3\times {10}^6$electron. One proton has a charge of $e=-1.6\times {10}^{-19}\mathrm{C}$.

So, to calculate the flux, we take the charges inside the cylinder, which is the negative charge $-1nC$, while the values outside the Gaussian surface are $+100nC$and $-100nC$. The inert charge is a charge that has no effect on the body.

localid="1649247619373" $\begin{array}{r}{Q}_{\text{in}}=ne\end{array}$

localid="1649247744113" $=\left(55.3\times {10}^6\right)\left(-1.6\times {10}^{-19}\mathrm{C}\right)$

localid="1649248372694" $\begin{array}{r}=-8.85\times {10}^{-12}\mathrm{C}\end{array}$

To get $\begin{array}{r}{\mathrm{\Phi }}_{\mathrm{e}}\end{array}$, we insert the values for $Q_{\text{in}}\text{and}{ϵ}_o=8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}\cdot {\mathrm{m}}^2$into equation ($1$).

localid="1649248364181" $\begin{array}{r}{\mathrm{\Phi }}_{\mathrm{e}}=\frac{Q_{\mathrm{in}}}{{ϵ}_o}\end{array}$

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

$=-1\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$