If electric flux is coming out of a Gaussian | vedclass

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(B) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q_{	ext{enc}}}{\varepsilon_0}$,where $Q_{	e

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Positive charge

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(B) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q_{	ext{enc}}}{\varepsilon_0}$,where $Q_{	ext{enc}}$ is the net charge enclosed by the surface.If the electric flux is coming out of the surface (outward flux),the value of $\phi$ is positive.Since $\varepsilon_0$ is always positive,for $\phi$ to be positive,the net enclosed charge $Q_{	ext{enc}}$ must be positive.Therefore,a positive charge is associated with the surface.

If electric flux is coming out of a Gaussian surface,what is associated with the surface?

Similar Questions

Give characteristics of electric field lines.

$A$ square surface of side $L$ meters is in the plane of the paper. $A$ uniform electric field $\vec{E} \text{ (V/m)}$ is also in the plane of the paper,limited only to the lower half of the square surface. The electric flux associated with the surface in $SI$ units is:

The number of electric lines that emerge from a finite charge $+q$ is . . . . . . .

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Reason: Electric field at all points on the Gaussian surface depends only on charges $q_1$ and $q_2$.

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