Question

If the electric flux entering and leaving an enclosed surface area are ' \(\phi_1\) ' and ' \(\phi_2\) ' respectively, the electric charge inside the surface will be ( \(\varepsilon_0=\) permittivity of free space)

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

Hint:

The net outward flux is the leaving flux minus the entering flux, and this equals charge over epsilon-zero.

Answer 1

The Correct Option is B

Step 1: Sign Convention
Outward flux is counted positive and inward flux negative.

Step 2: Apply
$\oint \vec E\cdot d\vec A = \phi_2 - \phi_1 = \dfrac{q}{\varepsilon_0}$.

Step 3: Result
$q = \varepsilon_0(\phi_2 - \phi_1)$, option B.

Final Answer:
q equals eps0 (phi2 - phi1). \[ \boxed{\text{(B) }\epsilon_{0}(\phi_{2}-\phi_{1})} \]