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

F = 0.257 N

Step-by-step explanation:

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

`F = (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 = ((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