Gauss's law states that

  • A
    the total electric flux through a closed surface is $\frac{1}{\varepsilon_0}$ times the total charge placed near the closed surface.
  • B
    the total electric flux through a closed surface is $\frac{1}{\varepsilon_0}$ times the total charge enclosed by the closed surface.
  • C
    the total electric flux through an open surface is $\frac{1}{\varepsilon_0}$ times the total charge placed near the open surface.
  • D
    the line integral of electric field around the boundary of an open surface is $\frac{1}{\varepsilon_0}$ times the total charge placed near the open surface.

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Similar Questions

An electric field is uniform,and in the positive $x$ direction for positive $x,$ and uniform with the same magnitude but in the negative $x$ direction for negative $x$. It is given that $E = 200 \hat{i} \; N/C$ for $x > 0$ and $E = -200 \hat{i} \; N/C$ for $x < 0$. $A$ right circular cylinder of length $20 \; cm$ and radius $5 \; cm$ has its centre at the origin and its axis along the $x$-axis so that one face is at $x = +10 \; cm$ and the other is at $x = -10 \; cm$.
$(a)$ What is the net outward flux through each flat face?
$(b)$ What is the flux through the side of the cylinder?
$(c)$ What is the net outward flux through the cylinder?
$(d)$ What is the net charge inside the cylinder?

The figure shows four charges $q_1, q_2, q_3$ and $q_4$ fixed in space. The total flux of the electric field through a closed surface $S$,due to all charges $q_1, q_2, q_3$ and $q_4$ is:

$A$ linear charge having linear charge density $\lambda$ penetrates a cube diagonally and then it penetrates a sphere diametrically as shown. What will be the ratio of flux coming out of the cube and the sphere?

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It is not convenient to use a spherical Gaussian surface to find the electric field due to an electric dipole using Gauss's theorem because

Is electric flux a scalar or a vector quantity?

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