Question

The electric flux entering and leaving an enclosed surface are \(\Phi _{1}\) and \(\Phi _{2}\) respectively. The electric charge inside the surface will be (\(\varepsilon _{0}\) = permittivity of vacuum)

• A. \(\frac{\phi _{2}-\phi _{1}}{\varepsilon _{0}}\)
• B. \(\frac{\phi _{1}+\phi _{2}}{\varepsilon _{0}}\)
• C. \(\varepsilon _{0}(\phi _{1}-\phi _{2})\)
• D. \(\varepsilon _{0}(\phi _{2}-\phi _{1})\)

Hint:

The net outward flux is the leaving flux minus the entering flux, and Gauss's law relates it to the enclosed charge.

Answer 1

The Correct Option is D

Step 1: Sign Convention
Outward flux is positive and inward flux is negative, so total flux is $\Phi_{2}-\Phi_{1}$.

Step 2: Apply the Law
$\oint\vec{E}\cdot d\vec{A}=\dfrac{q}{\varepsilon_{0}}$

Step 3: Result
$q=\varepsilon_{0}(\Phi_{2}-\Phi_{1})$, option (D).

Final Answer:
Outward minus inward flux times $\varepsilon_{0}$ gives the charge. \[ \boxed{\text{(D) }\varepsilon_{0}(\phi_{2}-\phi_{1})} \]