91Ó°ÊÓ

The term electric flux may be defined as the application of electric field that may be thought of as the number of electric lines of force that intersect a given area.

The total electric flux can be calculated by the relation

${\mathrm{F}}_{\mathrm{E}}=\frac{{\mathrm{Q}}_{\mathrm{end}}}{{\mathrm{Î\mu }}_0}$

Where role="math" localid="1664202829532" ${\mathrm{Q}}_{\mathrm{encl}}{\mathrm{Î\mu }}_0$total charged enclosed and electric constant respectively.

For radius 0.500 m.There is no any charge in the radius 0.500m so electric flux is zero.

For radius 1.50m.contain${\mathrm{q}}_2=-6.00\mathrm{nC}$ so the electric flux can be calculated as

$$\begin{gathered} {\mathrm{F}}_{\mathrm{E}}=\frac{-6.00×{10}^{-9}\mathrm{C}}{8.85×{10}^{-12}{\mathrm{C}}^{12}/\mathrm{N}.{\mathrm{m}}^2} \\ {\mathrm{F}}_{\mathrm{E}}=-678.{10}^{-12}\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the total electric flux due to these two-point charges through a spherical surface centered at the origin and with radius 0.500m is$0\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.

For radius 2.50m contain ${\mathrm{q}}_1=4.00\mathrm{nC},{\mathrm{q}}_2=-6.00\mathrm{nC}$so total charged enclosed is

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{end}}=\left(4.00-6.00\right)\mathrm{nC} \\ {\mathrm{Q}}_{\mathrm{encl}}=-2.00\mathrm{nC} \end{gathered}$$

Hence, the total electric flux due to these two-point charges through a spherical surface centered at the origin and with radius 1.50m is$-6.78.0{\mathrm{Nm}}^2/\mathrm{C}$.

Now the electric flux can be calculated as

$$\begin{gathered} {\mathrm{F}}_{\mathrm{E}}=\frac{-2.00×{10}^{-9}\mathrm{C}}{8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2} \\ {\mathrm{F}}_{\mathrm{E}}=-226.0\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2 \end{gathered}$$

Hence, the total electric flux due to these two-point charges through a spherical surface centered at the origin and with radius 2.50m is$-226.0{\mathrm{Nm}}^2/\mathrm{C}$.