A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is

0.050 m away. What is the magnitude of the second charge?

Explanation:

According to columbs law;

F =

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

what was the answer?

laticia draws the diagram below to show earth’s magnetic field.

answer

the arrows to the right of earth should point up.