Two equal point charges of +3.00 * 10^ -6 C are placed 0.200 m apart. What are the magnitude and direction of the force each charge exerts on the other?

Answer 1

Here is your hint: Coulomb's law.

Answer 2

find the force exerted by these two equal charges by using
f=(1/4*pi*epsilon)*(q1*q2)/r^2
if f1 is the force exerted by one charge q1 and f2 be the force exerted by another charge q2 which are equal in their magnitude(as you've mentioned) then f1=f2=f.
i.e, f1 and f2 are equal in magnitude.
to find direction, use
tan(theta)=[q2*sin(alpha)/{q1+q2*cos(alpha)}]
where alpha is the angle between the vectors(forces) exerted by two charges.here alpha should be equal to 180 degree.

Answer 3

F=(k*q1*q2)/r^2
q1 and q2 are 3.00 * 10^ -6 C
r=0.200 m
k=9*10^9

Answer 4

The answer is 2.03 N using anonymous post and puling in the number. They repel each other because they have the like charges.