Question 822: Electrostats & Capacitance

Question

If the electric flux entering and leaving an enclosed surface are $\phi_1$ and $\phi_2$ respectively, the electric charge inside the surface will be

Accepted Answer

We can solve this using Gauss's Law, which states that:
$$\oint \mathbf{E}\cdot d\mathbf{A} = \frac{q_{	ext{enc}}}{\varepsilon_0}$$
Here's a step-by-step explanation:
The flux leaving the surface contributes positively with a value of $$\phi_2,$$ while the flux entering the surface contributes negatively with a value of $$\phi_1.$$ Thus, the net electric flux is:
$$\phi_{	ext{net}} = \phi_2 - \phi_1$$
Applying Gauss's Law:
$$\phi_{	ext{net}} = \frac{q_{	ext{enc}}}{\varepsilon_0}$$
Substituting for the net flux:
$$\phi_2 - \phi_1 = \frac{q_{	ext{enc}}}{\varepsilon_0}$$
Solving for the enclosed charge $$q_{	ext{enc}}$$:
$$q_{	ext{enc}} = \varepsilon_0 \left(\phi_2 - \phi_1ight)$$
Thus, the correct answer is:
Option D
$$\varepsilon_0 (\phi_2 - \phi_1)$$

Answer

$\frac{\phi_2-\phi_1}{\varepsilon_0}$

Answer

$\frac{\phi_1+\phi_2}{\varepsilon_0}$

Answer

$\varepsilon_0\left(\phi_1-\phi_2\right)$

Answer

$\varepsilon_0\left(\phi_2-\phi_1\right)$

Electrostats & Capacitance

If the electric flux entering and leaving an enclosed surface are \phi_1 and \phi_2 respectively, the electric charge inside the surface will be

Choose the correct answer: