Two particles with positive charges q_1= 0.540 nC and q_2= 8.20 nC are separated by a distance of 1.30 m.
Along the line connecting the two charges, at what distance from the charge q_1 is the total electric field from the two charges zero?
E = k*q/r^2
d = [-b +/ sqrt(b^2 - 4ac)]/[2a]
The Attempt at a Solution
I tried drawing a picture:
q_1 (0.540 nC) O____>____.____<____O q_2 (8.20 nC)
O = charged particle
. = field point
The arrows show the direction of the electric fields. (Correct????)
x+y = 1.30 m
y = 1.30 m – x
E = (k*q_1)/(x^2) – (k*q_2)/(y^2)
0 = (k*q_1)/(x^2) – (k*q_2)/(y^2)
0 = (k*q_1)/(x^2) – (k*q_2)/(1.30-x)^2
0 = (k*q_1)/(x^2) – (k*q_2)/(x^2-2.60x+1.69)
(k*q_2)/(x^2-2.60x+1.69) = (k*q_1)/(x^2)
x^2*(k*q_2) – k*q_1(x^2-2.60x+1.69) = 0
x^2*(k*q_2) – (k*q_1*x^2) + (2.60*x*k*q_1) – 1.69*k*q_1 = 0
68.848x^2 + k*q_1*(2.60x –1.69) = 0
Using quadratic equation, x = 0.265 m or –0.449 m
Distance from q_1 = 0.265 m????????