Question

A charge of -8.08 nC exerts an attractive force of 17.98 N on a 4.52 µC charge. What is the distance between the charges?

Answer 1

Distance between the charges which will be given as -

According to coulomb's law, we have

F = k |q₁| |q₂| / r²

where, k = proportionality constant = 9 x 10⁹ Nm²/C²

F = attractive force = 17.98 N

q₁ = charge on first particle = - 8.08 x 10^-9 C

q₂ = charge on second particle = 4.52 x 10^-6 C

then, we get

(17.98 N) = (9 x 10⁹ Nm²/C²) |(-8.08 x 10^-9 C)| |(4.52 x 10^-6 C)| / r²

r² = [(3.28 x 10^-4 Nm²) / (17.98 N)]

r = 1.82 x 10^-5 m²

r = 0.00426 m

converting m into cm :

r = 0.426 cm