Physics, 20.08.2019 01:30 halimomohamed

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?

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Answer from: girlieredc

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

Answer from: Quest

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Answer from: Quest

answer and explanation;

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Two particles with charges of 5.00 μ c and -3.00 μc are placed 0.250 m apart. where can a third char...

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