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જો બંધ પૃષ્ઠમાં દાખલ થતું અને બહાર આવતું ફલક્સ અનુક્રમે $\phi_1$ અને $\phi_2$ છે. પૃષ્ઠની અંદરની બાજુએ વિદ્યુતભાર ........ હશે.
$({\phi _1}\, + \,\,{\phi _2}){\varepsilon _0}$
$({\phi _2}\, - \,\,{\phi _1}){\varepsilon _0}$
$\frac{{{\phi _1}\, + \,\,{\phi _2}}}{{{\varepsilon _0}}}$
$\frac{{{\phi _2}\, - \,\,{\phi _1}}}{{{\varepsilon _0}}}$
Solution
Hint: Here, we have to apply gauss's law to find electric charge.
Solution:
We know that, electric flux $\phi_1$ (or electric field lines) entering in a closed
surface is -ve and electric flux $\phi_2$ (or electric field lines) leaving a closed surface is $+ve.$
Hence, net electric flux through the closed surface,
$\phi=\phi_2-\phi_1$
Now, according to Gauss' theorem, the net electric flux $\phi$ passing through a closed surface is equal to the $1 / \varepsilon_0$ times of the total charge $q$, inside the surface.
Step1: Apply gauss's law
Given, Net electric flux, $\phi=\left(\phi_2-\phi_1\right)$
$\phi=\frac{ q _{\text {in }}}{\varepsilon_0}$
$\Rightarrow q_{\text {in }}=\varepsilon_0 \phi$
$\therefore q_{\text {in }}=\left(\phi_2-\phi_1\right) \varepsilon_0$
