Coulomb's law?

Question

A 2.2 times 10 to the negative 9 C charge is on the x axis at x= -1.5m and a 5.4 times 10 to the negative 9 C charge is on the x axis x=2m. Find the net force exerted on a 3.5 times 10 to the negative 9 C charge located at the origin.

Answer 1

This is easy - the charges all lie on the same axis, so you don't have to worry about vectors (only the x vector needs to be worried about, so treat this as a scalar problem).

The net force exerted on the charge at the origin can be found by the summation of the charges acting upon it at the origin. Coulomb's law states that:

F= (|q1||q2|)/(4*pi*epsilon_0*r^2)

Where q1 and q2 are the abosulte value of the two charges in coulombs, epsilon_0 is the permittivity of free space and is equal to 8.85x10^-12 F/m, and r is the distance between the two charges.

So since there are 2 charges acting on the charge at the origin, you will need to use coulomb's law twice, using the charge at the origin as one of the two q's in the equation both times. Just make sure you use the same charge value as the distance it is from the origin.

Let's let F1= the force on the charge at the origin from the charge located at 2m and F2=the force on the charge at the origin from the charge located at -1.5m. The total net charge, F, can then be stated as follows:

F=F1-F2

Answer 2

To do this problem, determine the forces separately and then add them.

k = coulomb's constant = 9 X 10 ^9 Nm^2/C^2
F= k (Q1)(Q2)/r^2

F1= (9 X 10^9) (2.2 X 10^ -9)(3.5 X 10^-9)/ (1.5^2)

F2 = (9 X10^9) (5.4 X 10^ -9) (3.5 X 10^ -9) / ( 2^2)

F1 = 3.08 X 10^ -8 Newtons
F2 = 4.24 X 10 ^ -8 Newtons

Since F1 points right and F2 points left :
Net force = F1 + (-F2) = 1.16 X 10^ - 8 Newtons left

Answer 3

10.2