engineering math help. electromagnetics...?

Question

please provide a solution for this problem...

let a point charge Q1=25nC(nano Coulumb) be located at P1(4,-2,7) and a charge Q2=60nC be at P2(-3,4,-2).

(a.)Find E at P3(1,2,3) ---> the answer here is 4.58ax - 0.15ay +5.51 az

(b.)At what point in the y-axis is Ex=0 ---> the answer here is -6.89 or -22.11

ive already solved the solution for letter A,,, but i dont know how to get letter B.. please give me a solution that will give that answer.. please write the steps.... thanks soooo much....

Answer 1

Vector difference P1 to (0, y, 0) is (-4, y + 2, -7).
Squared distance is 16 + y^2 + 4y + 4 + 49 = y^2 + 4y + 69

Vector difference P2 to (0, y, 0) is (3, y - 4, 2).
Squared distance is 9 + y^2 - 8y + 16 + 4 = y^2 - 8y + 29

So you want the x component of the field, which is the appropriate constant times:

25 * -4 * (y^2 + 4y + 69)^(-3/2) + 60 * 3 * (y^2 - 8y + 29)^(-3/2)

to be 0. Therefore

60 * 3 * (y^2 - 8y + 29)^(-3/2) = 25 * 4 * (y^2 + 4y + 69)^(-3/2)

Square both sides:

180^2 / (y^2 - 8y + 29)^3 = 100^2 / (y^2 + 4y + 69)^3

Multiply through by both denominators:

180^2 (y^2 + 4y + 69)^3 = 100^2 (y^2 - 8y + 29)^3

Numerically find the roots of left hand side minus right hand side equals zero. There are sign changes pos. to neg. between -23 and -22 and neg. to pos. between -7 and -6.

Dan