Question

The net electric flux through an octahedron is $-1000{\mathrm{Nmm}}^2/\mathrm{C}$. How much charge is enclosed within the octahedron?

Short answer

Charge enclosed with octahedron$Q_{\text{in}}=-8.85\mathrm{nC}$

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="1649251550682" ${\mathrm{\Phi }}_{\mathrm{e}}=\text{∮}\overrightarrow{E}\text{X}d\overrightarrow{A}=\frac{Q_{\text{in}}}{{ϵ}_o}$

The electric flow is determined by the charge inside the circular area, as indicated. There is no flux owing to charges outside the closed path.

Calculate equation ($1$) for $Q_{\mathrm{in}}$, which is the form we want to be in.

Equation $2$

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

We next input the values for ${\mathrm{\Phi }}_{\mathrm{e}}\text{and}{ϵ}_o$into equation ($2$) to obtain

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

$\begin{array}{r}=\left(-1000\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}\right)\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}\cdot {\mathrm{m}}^2\right)\\ \end{array}$

$=-8.85\times {10}^{-9}\mathrm{C}$

The charge in nC is

$$\begin{gathered} Q_{\text{in}}=\left(-8.85\times {10}^{-9}\mathrm{C}\right)\left(\frac{1\mathrm{nC}}{1\times {10}^{-9}\mathrm{C}}\right) \\ =-8.85\mathrm{nC} \end{gathered}$$