Question 818: Electrostats & Capacitance

Question

If the electric flux entering and leaving an enclosed surface is $\phi_1$ and $\phi_2$ then charge enclosed in the surface is ($\varepsilon_0=$ permittivity of free space)

Accepted Answer

Gauss's law tells us that the net electric flux through a closed surface is related to the charge enclosed by the surface. Specifically, GaussÃ¢ÂÂs law is:
$$ \oint_S \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{	ext{enc}}}{\varepsilon_0} $$
HereÃ¢ÂÂs how to apply this to the problem:
The net electric flux through the surface is the algebraic sum of the flux leaving the surface and the flux entering the surface. If you take $\phi_2$ as the flux exiting the surface and $\phi_1$ as the flux entering, then the net flux is:
$$    \Phi_{	ext{net}} = \phi_2 - \phi_1    $$
By substituting this into GaussÃ¢ÂÂs law, we have:
$$    \phi_2 - \phi_1 = \frac{Q_{	ext{enc}}}{\varepsilon_0}    $$
Solving for the enclosed charge $Q_{	ext{enc}}$, we multiply both sides by $\varepsilon_0$:
$$    Q_{	ext{enc}} = \varepsilon_0 (\phi_2 - \phi_1)    $$
Thus, the correct answer is:
Option D: $$\varepsilon_0\left(\phi_2 - \phi_1ight)$$

Answer

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

Answer

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

Answer

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

Answer

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

Electrostats & Capacitance

If the electric flux entering and leaving an enclosed surface is \phi_1 and \phi_2 then charge enclosed in the surface is (\varepsilon_0= permittivity of free space)

Choose the correct answer: