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A charge Q is enclosed by a Gaussian sphericalsurface of radius R. If the radius is doubled, thenthe outward electric flux will [2011]
- a)increase four times
- b)be reduced to half
- c)remain the same
- d)be doubled
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A charge Q is enclosed by a Gaussian sphericalsurface of radius R. If ...
By Gauss’s theorem,
Thus, the net flux depends only on the
charge enclosed by the surface. Hence,
there will be no effect on the net flux if the
radius of the surface is doubled
charge enclosed by the surface. Hence,
there will be no effect on the net flux if the
radius of the surface is doubled
Most Upvoted Answer
A charge Q is enclosed by a Gaussian sphericalsurface of radius R. If ...
Explanation:
The electric flux through a Gaussian surface is given by the formula:
Φ = Q/ε0
Where, Q is the charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.
When the radius of the Gaussian surface is doubled, its area becomes four times the original area (since the surface area of a sphere is proportional to the square of its radius).
But the charge enclosed by the Gaussian surface remains the same.
Therefore, the electric field at any point on the Gaussian surface will be the same as before (since the electric field due to a point charge varies inversely with the square of the distance from the charge).
Hence, the electric flux through the Gaussian surface will also remain the same.
Therefore, the correct answer is option 'C': Remain the same.
The electric flux through a Gaussian surface is given by the formula:
Φ = Q/ε0
Where, Q is the charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.
When the radius of the Gaussian surface is doubled, its area becomes four times the original area (since the surface area of a sphere is proportional to the square of its radius).
But the charge enclosed by the Gaussian surface remains the same.
Therefore, the electric field at any point on the Gaussian surface will be the same as before (since the electric field due to a point charge varies inversely with the square of the distance from the charge).
Hence, the electric flux through the Gaussian surface will also remain the same.
Therefore, the correct answer is option 'C': Remain the same.
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