Question

Two point charges, 3.0 µC and -2.0 µC, are placed 4.8 cm apart on the x axis, such that the -2.0 µC charge is at x = 0 and the 3.0 µC charge is at x = 4.8 cm. At what point(s) along the x axis is the potential zero?
I know that between the two charges, the answer is 1.92 cm by solving this equation -2.0µC(x-0.048m)^2 = 3.0µC(x^2) for x, and I know that when x > 4.8, there is no point where the potential is zero. I found out that when x < 0, the answer is -9.6, but I don't know how to arrive at that answer. If anyone can show me the process to getting that answer, I'd be very appreciative.

Answer 1

Never mind. I figured it out.

Answer 2

I asks where the potential is zero, not necessarily where the force is zero

You can add the potential due to the two charges on the x axis

V = k Q/R1 = k Q1 /|x-xq|
forget k and the 10^-6 because they are the same for both and we are looking for zero

I will do the - 2 first
-2/|x|
then the +3 charge
+3/|x-4.8|
where will the sum be 0 ?
look at negative x first
d is distance left of x = 0
-2/d +3/(4.8+d)=0
3 d = 9.6 + 2 d
d = 9.6 left of zero
x = -9.6 (The only place)
look between x = 0 and x = 4.8
d is now right of x = 0
-2/d + 3/(4.8-d) = 0
3 d =+9.6 -2d
d = 9.6 nope, x not between the two

look right of x = 4.8, d is x value must be >4.8
-2/d + 3/(d-4.8) = 0
3 d = -2 d + 9.6
5 d = 9.6
d = 1.92 Nope, not to right of 4.8
The only place where there can be no potential is on the opposite side of the 2 charge from the 3 charge, negative x