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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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Answer» According to Gauss's law, flux through a closed surface is given by, \(\phi = \frac{q}{\in_o}\) Here, q is the charge enclosed by the gaussian surface. That is, on increasing the radius of the gaussian surface, charge q remains unchanged. So, flux through the gaussian surface will not be affected when its radius is increased. |
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