A) zero
B) half that due to a single charge
C) double that due to a single charge
D) dependent on the position of the dipole
Correct Answer: A
Solution :
Total electric flux through enclosed surface\[\phi =\frac{q}{{{\varepsilon }_{0}}}+\left( \frac{-q}{{{\varepsilon }_{0}}} \right)=0\] Note: As Gaussian surface is not a enclosed surface, hence the question should have been an enclosed surface and not a Gaussian surface.
You need to login to perform this action.
You will be redirected in
3 sec
