Two objects are charged and their centers are separated by 28.6 cm. Object A is charged to 34 nC and object B is charged to 38 nC. What is the magnitude of the electrostatic force between them?
- electron1Lv 710 months agoFavorite Answer
The following equation is used to solve this problem.
F = k * Q1 * Q2 ÷ d^2
I use 9 * 10^9 for k. The unit of the charge must be in coulombs, and the unit of distance must be in meters.
Q1 = 3.4 * 10^-8 C
Q2 = 3.8 * 10^-8 C
d = 0.286 m
F = 9 * 10^9 * 3.4 * 10^-8 * 3.8 * 10^-8 ÷ 0.286^2 = 1.1628 * 10^-5 ÷ 8.1796 * 10^-2
The force is approximately 1.42 * 10^-4 N.
- billrussell42Lv 710 months ago
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
F = (9e9)(34e-9)(38e-9) / 0.286²
you can do the math.