- #1
marissag
- 11
- 0
Homework Statement
A positive point charge Q1 = 1.9 x 10^-5C is fixed at the origin of coordinates, and aa negative charge Q2 = -6.5 x 10^-6C is fixed to the axis at x = +2.0m. Find the location of the place(s) along the x-axis where the electric field due to these two charges is zero. IF there is only one location then enter zero for the second location.
x = ______m
x = ______m
The Attempt at a Solution
for the distance I used Q1 = x and Q2 = abs val of 2-x and then used the equation of Q1/r1^2 = Q2/r2^2
so I started out with ::::
first (1.9 x 10^-5C)/x^2 = (-6.5 x 10^-6C)/((2-x)^2)
then (1.9 x 10^-5C)/x^2 = (-6.5 x 10^-6C)/(x^2-4x+4)
then multiplied both denominators to get rid of them and got:::
(1.9 x 10^-5C)(x^2-4x+4) = (-6.5 x 10^-6C)(x^2)
then multiplied them all out and got:::
(2.55 x 10^-5C)(x2) - (7.6 x 10^-5)(x) + ((7.6 x 10^-5) = 0
at this point I used the quadratic formula and used:::
a = 2.55 x 10^-5
b = 7.5 x 10^-5
c = 7.5 x 10^-5
Now...here is the problem. When I plug these all into the quadratic formula I end up with a negative number in the square root which you can't take the square root of...
any help is appreciated...anybody to show me where I went wrong is appreciated as well!