flux through a gaussian surface which has enclosed a dipole ??

Asked by kothari_prachi | 25th May, 2010, 11:12: AM

Expert Answer:

Dear student,

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

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.