Question

A positive charge q₁ = 1.15 μC is fixed at the origin, and a second charge q₂ = -2.59 μC is fixed at x = 15.9 cm. Where along thex-axis should a third charge be positioned so that it experiences no force?

Answer 1

Lets assume that 3rd charge is placed at a distance xcm from origin.

q1 = 1.15*10^-6 C
r13 = (x-0) = x cm = x*10^-2 m

q2 = -2.59*10^-6 C
r23 = (x-15.9)*10^-2 m

Net electric field at position of 3 = K*(q1/r12^2 + q2/r23^2)
= (9*10^9)* (1.15*10^-6/ (x*10^-2)^2 + -2.59*10^-6 / ((x-15.9)*10^-2)^2 )
= (9*10^13)* (1.15*10^-6/ x^2 -2.59*10^-6 / (x-15.9)^2)
This electric field should be 0 for net Force to be 0
(9*10^13)* (1.15*10^-6/ x^2 -2.59*10^-6 / (x-15.9)^2) = 0
1.15*10^-6/ x^2 - 2.59*10^-6 / (x-15.9)^2 = 0
1.15*10^-6/ x^2 = 2.59*10^-6 / (x-15.9)^2
2.25 x^2 = (x-15.9)^2
2.25 x^2 = x^2 - 31.8x + 252.81
1.25 x^2 + 31.8x -252.81 =0
solving we get x = -31.8 cm and 6.4 cm