An electric dipole is an object with a net charge of zero an

An electric dipole is an object with a net charge of zero and characterized by a separation of positive and negative charges. We can model an electric dipole as made of two equal and opposite point charges of magnitude Q (so that one point charge has charge +Q and the other has charge-0) separated by a constant distance d. For example, we can build a dipole by putting the charges on the weights at the end of a dumbbell. is at coordinates (x, y, z) Consider the dipole shown below (not to scale!), oriented so that the positive charge (+d2, 0, 0) and the negative charge-Q is at coordinates (x, y, (-d 2,0, o). 10d -Q t0 a) [3 points] A positive charge of magnitude +q (that is, a th a) 13 points] A positive charge of magnitude +q (that is, a third, diferent charge) is placed to the right of the pole at the point P shown in the picture above. The position of point P is (10d, 0, o). Find the net electric multiple force (magnitude and direction) exerted on the charge ta due to the dipole. Express your answer as a of the quantity kaOld. (Hint: Use superposition! First find the force on the +q charge due to the +0 charge then find the force on the tq charge due to the -Q charge.] FC45a) b) [2 points] What is the electric field (magnitude and direction) due to the dipole at point P Express your part (a) to make this part answer as a multiple of the quantity kOd. [Hint: You can use your answer from really easy [This part is a litle more imolved computationalily and will not count towards your grade on this assignment I am including it for you to think about and try We will go over it at the beginning of lecture next week] Consider the point R on the ty-axis shown in the picture above. The position of point R is (0,y, O). Find the net electric field at point R (magnitude and direction). 2KG

Solution

á)

The froce b by +Q on +q F1 = +kQq / 9.5d^2 in +ve x direction     , distance = 10d-d/2 =9.5d

The froce by -Q and +q F2 = -kQq / 10.5d^2    in -x direciton distance = 10d+d/2 = 10.5d

net force on q due to +Q and -Q is Fnet = [kQq / (9.5d)^2] -[kQq / (10.5d)^2] = 0.002 * kQq / d^2

b)

Electric field at P ,E = Fnet /q = 0.002 * kQ / d^2 in +ve x direcion.

c)

distance from +Q to R ; d1 = sqrt (y^2+d^2/4) similalrly

distance from -Q to R ; d2 = sqrt (y^2+d^2/4) similalrly

Enet = kQ / d1^2- kQ / d2^2 = 0             since(d1=d2)