Thank you for reporting, we will resolve it shortly
Q.
What is the electric flux linked with closed surface ?
Electric Charges and Fields
Solution:
According to Gauss’s law
Electric flux, $\phi=\frac{q}{\varepsilon_{0}}$
where, q = total charge enclosed by closed surface
$\therefore \, \phi=\frac{1.25+7+1-0.4}{\varepsilon_{0}}
=\frac{8.85 C}{8.85\times10 ^{-12} C^{2}N^{-1}m^{-2}}=10^{12}N m^{2} C^{-1}$
