Question

Both in class and in Knight (p. 633) it was shown that the electric field of an electric dipole far along the dipole axis is given by 2kp In the second problem of PS2 (Knight Chapter 23, Problem 3) you determined the electric field of a specific electric dipole at a specified point along the bisecting plane of the dipole. Repeat this calculation for an electric dipole with an unspecified charge q and show that the electric field very far from the dipole can be approximated as given by equation 23.11 in Knight: kp bp 3 Note that quantities in boldface are vector quantities.

Answer 1

Electric field by dipole at bisected point which is very far from it

Electric field by dipole at bisector Electric field at point p y + 9Vertical-bar E + Vertical-bar = k 9/(squareroot r^2 + l^2)^2 and electric field at point p by minus 9 E_minus = k9/(squareroot r^2 + l^2)^2 then Vertical-bar E_+ Vertical-bar = Vertical-bar E_minus Vertical-bar so vertical component cancelled each other and E_p = E + cos theta + E minus cos theta \r\n E_p = 2 k 9/(squareroot r^2 + l^2)^2 cos theta (minus l) from figure cos theta = l/ squareroot r^2 + l^2 hence E_p = 2 k9/(squareroot r^2 + l^2)^2 l/ squareroot r^2 + l^2 (minus i^^) E_p = 2 k9 l/(r^2 + l^2)^3/2 (minus i^^) for long distance r > > l, E_p = 2 k9 (minus l^^)/r^3 and p = 29 l (i^^), so E^rightarrow_p = minus k p^right arrow/r^3