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

Answer:

0.86 m

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

$\frac{kq_{2}q}{d^{2}} = \frac{kq_{1}q}{(r+d)^{2}}$

$\frac{q_{2}}{d^{2}} = \frac{q_{1}}{(r+d)^{2}}$

$\frac{(5\times 10^{-6})}{(0.250+d)^{2}} = \frac{(3\times 10^{-6})}{(d^{2}}$

d = 0.86 m