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Question
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?
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Solution
According to Gauss's law, flux through a closed surface is given by
`phi = q/ε_0`
Here, q is the charge enclosed by the Gaussian surface.
Since, on increasing the radius of the Gaussian surface, charge q remains unchanged, the flux through the spherical Gaussian surface will not be affected when its radius is increased.
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| (i) | (ii) | (iii) | (iv) |
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