A negative charge, q1, of 6 µC is 0.002 m north of a positive charge, q2, of 3 µC. What is the magnitude and direction of the electrical force, Fe, applied by q1 on q2?

A) magnitude: 8 * 10^1 N

direction: south

B) magnitude: 8 * 10^1 N

direction: north

C) magnitude: 4 * 10^4 N

direction: south

D) magnitude: 4 * 10^4 N

direction: north

D) magnitude: 4 * 10^4 N

direction: north

Explanation:

q₁ = magnitude of negative charge = 6 x 10⁻⁶ C

q₂ = magnitude of positive charge = 3 x 10⁻⁶ C

r = distance between the two charges = 0.002 m

Magnitude of electric force between the two charges is given as

F = k q₁ q₂/r²

where k = constant = 9 x 10⁹

inserting the values

F = (9 x 10⁹) (6 x 10⁻⁶) (3 x 10⁻⁶) / (0.002) ²

F = 4 x 10⁴ N

magnitude : 4 x 10⁴ N

Direction : North

we know that a negative charge pulls a positive charge towards it. the negative charge "q₁" is north of the positive charge. hence the charge q₁ apply force on charge q₂ in north direction.

D. magnitude: 4 * 104 N

direction: north