The inward and outward electric flux for a cl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) According to Gauss's law, the net electric flux $\phi_{net}$ through a closed surface is given by $\phi_{net} = \frac{Q_{enclosed}}{\varepsilon_0}

Vedclass · Accepted Answer

$-4 	imes 10^3 \varepsilon_0 	ext{ C}$

Vedclass · Answer

(D) According to Gauss's law, the net electric flux $\phi_{net}$ through a closed surface is given by $\phi_{net} = \frac{Q_{enclosed}}{\varepsilon_0}$.Inward flux is taken as negative and outward flux is taken as positive.Given: Inward flux $\phi_{in} = -8 	imes 10^3 	ext{ N} \cdot m^2/C$ and outward flux $\phi_{out} = 4 	imes 10^3 	ext{ N} \cdot m^2/C$.The net flux $\phi_{net} = \phi_{in} + \phi_{out} = -8 	imes 10^3 + 4 	imes 10^3 = -4 	imes 10^3 	ext{ N} \cdot m^2/C$.Using Gauss's law: $Q_{enclosed} = \phi_{net} 	imes \varepsilon_0$.Therefore, $Q_{enclosed} = (-4 	imes 10^3) \varepsilon_0 	ext{ C}$.

The inward and outward electric flux for a closed surface in units of $N \cdot m^2/C$ are respectively $8 \times 10^3$ and $4 \times 10^3$. Then the total charge inside the surface is [where $\varepsilon_0$ = permittivity constant].

Similar Questions

$Assertion (A):$ $A$ charge $q$ is placed at a height $h/4$ above the center of a square of side $b$. The flux associated with the square is independent of the side length $b$.
$Reason (R):$ Gauss's law is independent of the size of the Gaussian surface.

$A$ charge $Q$ is placed at the center of a closed cube. The flux passing through any one face of the cube will be .......

The figure shows the electric field lines of three charges with charges $+1, +1$,and $-1$. The Gaussian surface in the figure is a sphere containing two of the charges. The total electric flux through the spherical Gaussian surface is

Two concentric spherical surfaces $P_1$ and $P_2$ enclose charges $\frac{Q}{2}$ and $4Q$ as shown in the figure. If $\phi_1$ and $\phi_2$ are the electric fluxes linked with the surfaces $P_1$ and $P_2$ respectively,then:

The dimensional formula of electric flux is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module