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point charges q1=2.4 μC and q2=9.5 μC are brought near eachother each a repulsive force of 0.82 N.Whats the distance between the charges?
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- billrussell42Lv 73 years agoFavorite Answer
F = kQ₁Q₂/r²
(9e9)(2.4µ)(9.5µ) / r² = 0.82
r² = (9e9)(2.4µ)(9.5µ) / 0.82 = 250 e-3
r = 0.50 meter
Coulomb's law, force of attraction/repulsion
F = kQ₁Q₂/r²
Q₁ and Q₂ are the charges in coulombs
F is force in newtons
r is separation in meters
k = 8.99e9 Nm²/C²
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