Question

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?

Answer 1

D) magnitude: 4 × 104 N

direction: north

Step-by-step explanation:

on edg

Answer 2

Force,F =
`4.05* 10^4\ N`

Step-by-step explanation:

Given that,

Charge 1,
`q_1=-6\ \mu C=-6* 10^(-6)\ C`

Charge 2,
`q_2=+3\ \mu C=3* 10^(-6)\ C`

Negative charge is 0.002 m north of a positive charge, distance between charges, d = 0.002 m.

Let
`F_e` is the magnitude of electrical force applied by q₁ on q₂. We know that the electrical force is given by the formula as follows :

`F_e=k(q_1q_2)/(d^2)\\\\F_e=9* 10^9* (6* 10^(-6)* 3* 10^(-6))/((0.002)^2)\\\\F_e=4.05* 10^4\ N`

So, the electrical force applied by q₁ on q₂ is
`4.05* 10^4\ N` towards north.