flux through a gaussian surface which has enclosed a dipole ??
Asked by kothari_prachi | 25th May, 2010, 11:12: AM
We have an electric dipole. draw a Gaussian surface around our electric dipole. Now, the total charge enclosed by our Gaussian surface is zero, so according the Gauss' Law the flux through the Gaussian surface is zero, and so is the electric field intensity due the electric dipole.
Answered by | 25th May, 2010, 11:26: AM
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