"Can you arrange the two point charges q1 =−2.0×10−6 C and q2 =4.0×10−6 C along the x-axis so that E =0 at the origin?"
E = K (|q|/r2)
K = Coloumb Constant
The Attempt at a Solution
Since one particle is negatively charged and the other is positively charged, they'll both be pointing their fields down the same direction, just at different strengths. The distance between the two charges will be named with an r, and the distance between q1 to 0 will be x (meaning q2 distance to 0 is x+r)
0 = Eox = E2x + E1x
0 = Eox = K (|q2|/(x+r)2) + K (|q1|/x2)
0 = Eox = K (|q2|/(x+r)2 + |q1|/x2)
0 = |q2|/(x2(x+r)2) + |q1|/(x2(x+r)2
0 = |q2|/(x2 + |q1| (x+r)2)/(x2(x+r)2
0 = |q2|x2 + |q1| (x+r)2
Before I go any further, this doesn't look correct. Am I on the right path?