two point charges, q1 = +25 nc and q2 = −75 nc, are separated by a distance r = 3.0 cm. find the magnitude and direction of the electric force (a) that q1 exerts on q2 and (b) that q2 exerts on q1.
Answers
Given: Two point charges, q1 = +25 NC and q2 = −75 nc, the distance between them r = 3.0 cm
To find the magnitude and direction of the electric force (a) that q1 exerts on q2 and (b) that q2 exerts on q1.
Solution: According to coulomb's law
F= kq1q2/ r2
where q1 and q2 are charged and r is the distance between them.
putting the value of q1 q2 and r in the above equation,
F= (9×10^9)×(25×10^-9)×(-75×10^-9)/ (3×10^-2)^2
F= −0.01875 N
the magnitude of the force that q1 exerts on q2 and q2 exerts on q1 will be the same.
Also, we can see that both the charges are of the opposite nature that is one is positive and one is negative. Therefore we can say that there will an attractive force between them.
So the magnitude that q1 exerts on q2 and (b) that q2 exerts on q1 will be 0.01875N and the direction of the force will be attractive.