Question

A small cork with an excess charge of +6.0 µC is placed 0.95 m from another cork, which carries a charge of -4.3 µC. What is the magnitude of the electric force between them?

Answer 1

Answer:

F = 0.257 N

Explanation:

The magnitude of the electric force between the charges can be calculated by using Colomb's Law:

$F = \frac{kq_1q_2}{r^2}$

where,

F = Electric Force = ?

k = Colomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = charge on first cork = 6 μC = 6 x 10⁻⁶ C

q₂ = charge on second cork = 4.3 μC = 4.3 x 10⁻⁶ C

r = distance between corks = 0.95 m

Therefore,

$F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(6\ x\ 10^{-6}\ C)(4.3\ x\ ^{-6}\ C)}{(0.95\ m)^2}$

F = 0.257 N