Question

Two particles with charges of 5.00 μ C and -3.00 μC are placed 0.250 m apart. Where can a third charge be placed so that the net force on it is zero?

Answer 1

0.86 m

Step-by-step explanation:

q₁ = magnitude of positive charge = 5 x 10⁻⁶ C

q₂ = magnitude of negative charge = 3 x 10⁻⁶ C

r = distance between the two charges = 0.250 m

d = distance of the location of third charge from negative charge

q = magnitude of charge on third charge

Using equilibrium of electric force on third charge

`(kq_(2)q)/(d^(2)) = (kq_(1)q)/((r+d)^(2))`

`(q_(2))/(d^(2)) = (q_(1))/((r+d)^(2))`

`((5* 10^(-6)))/((0.250+d)^(2)) = ((3* 10^(-6)))/((d^(2))`

d = 0.86 m