**Forces in a Three-Charge System**

**Solution Below:**

|F| = K*|QQ’|/(d^2),

where K = 1/(4πε_{0}. ε_{0} = 8.854 * 10^-12 C^2/(N*m^2) is the permittivity of free space.

Consider two point charges located on the x axis: one charge, q1 = -11.5 nC, is located at x1 = -1.685 m; the second charge, q2 = 40.0 nC, is at the origin (x = 0.0000).

Part A

Your answer may be positive or negative, depending on the direction of the force.

Just find the effects of q1 and q2 separately:

For q1 on q3:

The distance between q1 and q3 is -1.685m – (-1.120m) = 0.565

F_{1,3} = K * (11.5 nC * 47.0 nC) / (0.565^2)

Since the charges are in nanocoulombs:

F_{1,3} = (1.15 * 10^-8 * 4.70 * 10^-8) / (0.565^2)

F_{1,3} = 1.69316 * 10^-15 K

You can use the formula given for K, or just know that K is the coulomb constant and has a value of 8.9875 * 10^9:

F_{1,3} = (1.69316 * 10^-15) * (8.9875 * 10^9)

F_{1,3} = 1.52173 * 10^-5 N

Now do the same thing for the effect of charge 2 on charge 3:

The distance between charge 2 and 3 is 1.120m, so:

F_{2,3} = K * (40.0 nC * 47.0 nC) / (1.120^2)

F_{2,3} = 1.3469786 * 10^-5 N

Now go back and look at the positions and signs of each charge:

Since q1 is negative and q3 is positive, the charges will attract each other. Q1 is to the left of q3, so q3 will move left (negative)

Since q2 and q3 are both positive, the charges will repel each other. Q2 is to the right of q3, so q3 will move left (also negative).

You can see that the effect of both q1 and q2 is to cause q3 to move left. So you add the forces from each together:

F = -1.3469786 * 10^-5 N – 1.52173 * 10^-5 N

F = -2.87 * 10^-5 N

F = -2.87 * 10^-5 N

Thanks for the great explanation!!

You’re very welcome!