The electric flux linked with the closed surface in $\mathrm{Nm}^2 \mathrm{C}^{-1}$ is…
The electric flux linked with the closed surface in $\mathrm{Nm}^2 \mathrm{C}^{-1}$ is
$\left(\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)$
$10^{12}$
$8.85 \times 10^{-13}$
$10^{10}$
$10^{11}$
Solution
Electric flux, $\phi=\frac{q}{\varepsilon_0}$
where $q=$ total charge enclosed by closed surface
$q=(2.35+5+2-0.5) C$
$q=(2.35+5+2-0.5) \mathrm{C}$
$\phi=\frac{q}{\varepsilon_0}=\frac{8.85 \mathrm{C}}{8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}}=10^{12} \mathrm{Nm}^2 / \mathrm{C}$
