(B) According to Gauss's Law,the net electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free

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$(\varphi _2 - \varphi _1)\varepsilon _0$

(B) According to Gauss's Law,the net electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space: $\Phi_{net} = \frac{Q_{enc}}{\varepsilon_0}$.Here,the flux entering the surface is $\varphi_1$ (negative flux) and the flux leaving the surface is $\varphi_2$ (positive flux).Therefore,the net flux is $\Phi_{net} = \varphi_2 - \varphi_1$.Substituting this into Gauss's Law: $\varphi_2 - \varphi_1 = \frac{Q_{enc}}{\varepsilon_0}$.Thus,the charge enclosed is $Q_{enc} = (\varphi_2 - \varphi_1)\varepsilon_0$.

Class 12 Physics Electric Charges and Fields Electric Field Lines, Electric Flux and Gauss's Law

Easy

If the electric flux entering and leaving an enclosed surface respectively is ${\varphi _1}$ and ${\varphi _2}$,the electric charge inside the surface will be:

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A
$(\varphi _1 + \varphi _2)\varepsilon _0$
B
$(\varphi _2 - \varphi _1)\varepsilon _0$
C
$(\varphi _1 + \varphi _2)/\varepsilon _0$
D
$(\varphi _2 - \varphi _1)/\varepsilon _0$

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Electric Charges and Fields

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Electric Field Lines, Electric Flux and Gauss's Law

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If the electric flux entering and leaving an enclosed surface respectively is ${\varphi _1}$ and ${\varphi _2}$,the electric charge inside the surface will be:

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In a uniform electric field,a cube of side $1 \ cm$ is placed. The total energy stored in the cube is $8.85 \ \mu J$. The electric field is parallel to four of the faces of the cube. The electric flux through any one of the remaining two faces is

$A$ closed surface passes through a spherical conductor as shown in the figure. If a negative charge $-Q$ is placed at point $P$,what will be the nature of the electric flux coming out of the closed surface?

$Assertion (A):$ $A$ 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.

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$Assertion (A):$ $A$ 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.