Question

4. Consider two thin charged rods of length L carrying uniformity distributed charges of +q and -q positioned end-to-end as shown in the figure below. (a) Calculate the electric field at point P, a distance x to right of where the rods meet. (b) Show that the approximate expression for the electric field at point P for x >> L looks like that of a dipole (Hint: Review Example 21.14 in your text). (c) What is the dipole moment of the charge distribution for x >> L and why does your answer make sense?

Answer 1

Given that a) Electric Field due to positively charged rod, E_+ = 1/4 pi elementof_0 + q/(x - L)^2 Electric field due to negatively charged rod, E_- = 1/4 pi elementof_0 (-q)/(x + L)^2 Net electric field, E = E_+ + E_- E = 1/4 pi elementof_0 q/(x - L)^2 -1/4 pi elementof_0 q/(x + L)^2 E = q/4 pi elementof_0 [(x + L)^2 - (x - L)^2/(x - L)^2 - (x + L)^2] = q 4 times L/4 pi elementof_0 (x^2 - L^2)^2 E = q 4 times L/4 pi elementof_0 (x^2 - L^1)^2 \r\n b) if x > > L, E = 1/4 pi elementof_0 q 4 times L/(x^2 - L^2)^2 We have dipole moment p = 2qL E = 1/4 pi elementof_0 2 px/(x^2 - L^2)^2 for x > > L, E = 1/4 pi elementof_0 2 px/x^4 = 2p/4 pi elementof_0 x^3 c) dipole moment, p = 2 qL E = 1/4 pi elementof_0 2 px/(x^2 - L^2)^2 = 2p/4 PI elementof_0 x^3 E = 2p/4 pi elementof_0 x^3