Question

A charge of -0.4 µC is located at the origin; a charge of 0.53 µC is located at x = 0.2 m, y = 0; a third charge Q is located at x = 0.32 m, y = 0. The force on the 0.53 µC charge is 4.3 N, directed in the positive x direction.

1. Determine the charge

2. With this configuration of three charges, where, along the x direction, is the electric field zero? xf=

3. x2=

Answer 1

Take 1/4 = k = 9 x 10⁹.
Force on 0.53 C is
Assume that Q is attracting 0.53 C, the proper sign will come from the equation.
1/4 [(0.4 C x 0.53 C) / 0.2²] (-) + 1/4 [(Q x 0.53 C) / 0.12²] () = 4.3 ()
9 x 10⁹ x 0.53 x 10^-6 [-10 C+ 69.4 Q] = 4.3
69.4 Q - 10 C = 0.9 x 10^-3
Q = - 0.144 C (sign is negative since we assumed Q is attracting 0.53)

2.
Take that out test charge is at x m from the origin
The electric field at x is
-k[0.4 C / (x)²] + k[0.53 C / (x - 0.2)²] - k[0.144 C / (x - 0.32)²] = 0
-4 / (x)² + 5.3 / (x - 0.2)² - 1.44 / (x - 0.32)² = 0.
This is a 4th power polynomial and solving this will get the roots of x, which are the positions at electric field will be zero.