what is the magnitude and direction of electric field intensity due to an electric dipole of length 2a at the midpoint of the line joining with two charges?

1 Answer
Jan 3, 2018

#vec E = - (2kq)/a^2 haty #

Explanation:

#E = kq/r^2#

In the middle of the dipole, the electric fields contributed by both charges are in the same direction; they are both pointing toward the negative charge. Thus, the magnitude of the electric field is the sum of the contribution of the two charges.

#E = kq/a^2 + kq/a^2 = 2kq/a^2 #

The direction of the field is pointing to the negative charge. So its actual direction depends on how the dipole is oriented. If the dipole is oriented north-south and that the +q is the northern pole, then E is pointing to the south. That is,

#vec E = - (2kq)/a^2 haty #