Question

If the electric flux entering and leaving an enclosed surface respectively is $\phi_{1}$ and $\phi_{2}$, the electric charge inside the surface will be

• A. $\left(\phi_{2}-\phi_{1}\right)\varepsilon_{0}$
• B. $\frac{\left(\phi_{2}+\phi_{1}\right)}{\varepsilon_{0}}$
• C. $\frac{\left(\phi_{2}-\phi_{1}\right)}{\varepsilon_{0}}$
• D. $\left(\phi_{2}+\phi_{1}\right)\varepsilon_{0}$

Answer 1

Answer: (A)

According to Gauss’s theorem, charge in flux $\frac{charge\, enclosed\, by \,the\, surface}{\varepsilon_{0}}$
$\therefore q=\left(\phi_{2}-\phi_{1}\right)\varepsilon_{0}$.