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प्रश्न
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विकल्प
Gauss's Law is valid only for symmetrical charge distributions.
Gauss's Law is valid only for charges placed in vacuum.
The electric field calculated by Gauss's Law is the field due to the charge inside the Gaussian surface.
The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface.
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उत्तर
The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface.
The contribution of flux on the closed surface due to the charges lying outside the surface is zero because number of field line entering the closed surface is equal to the number of field lines coming out of the surface so the net contribution of the charge lying outside the closed surface to the flux is zero. Therefore, the net flux through the surface due to the charge lying outside the the closed surface is zero. The contribution that counts is only due to the charges lying within the closed surface.
Thus, the flux of the electric field through a closed surface due to all the charges (inside and outside the surface) is equal to the flux due to the charges enclosed by the surface
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संबंधित प्रश्न
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?
Write S.I unit of electric flux.
What is the electric flux through a cube of side 1 cm which encloses an electric dipole?
What is the net flux of the uniform electric field of previous question through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
Given a uniform electric filed \[\vec{E} = 4 \times {10}^3 \ \hat{i} N/C\]. Find the flux of this field through a square of 5 cm on a side whose plane is parallel to the Y-Z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis?
Two charges of magnitudes −3Q and + 2Q are located at points (a, 0) and (4a, 0) respectively. What is the electric flux due to these charges through a sphere of radius ‘5a’ with its centre at the origin?
A thin straight infinitely long conducting wire having charge density λ is enclosed by a cylindrical surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder.
If the flux of the electric field through a closed surface is zero,
(a) the electric field must be zero everywhere on the surface
(b) the electric field may be zero everywhere on the surface
(c) the charge inside the surface must be zero
(d) the charge in the vicinity of the surface must be zero
The following figure shows a closed surface that intersects a conducting sphere. If a positive charge is placed at point P, the flux of the electric field through the closed surface

The electric flux through a closed Gaussian surface depends upon ______.
If the flux associated with a coil changes at the rate of 360 webers every 4 minutes, then the induced e.m.f. is ______
The SI unit of electric flux is ______.
An electric charge q is placed at the centre of a cube of side a. The electric flux on one of its faces will be ______.
A point charge q is placed at a distance a/2 directly above the centre of a square of side a. The electric flux through the square is ______.
In a region of space having a uniform electric field E, a hemispherical bowl of radius r is placed. The electric flux Φ through the bowl is:
The electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10 cm surrounding the total charge is 20 V-m. The flux over a concentric sphere of radius 20 cm will be ______.
A hollow sphere of radius R has a point charge q at its centre. Electric flux emanating from the sphere is X. How will the electric flux change, if at all, when radius of the sphere is doubled?
A hollow cylinder has a charge of 'q' C within it. If 𝜙 is the electric flux associated with the curved surface B, the flux linked with the plane surface A will be ______.

