Point Charges in One Dimension

A point charge q1 = -3.5 μC is located at the origin of a co-ordinate system. Another point charge q2 = 5.1 μC is located along the x-axis at a distance x2 = 9.3 cm from q1. 1) What is F12,x, the value of the x-component of the force that q1 exerts on q2? -18.57

N

The basic equation for the Coulomb force is: A negative value of F means it’s an attractive force. You have to rely on context to know what direction the force acts in. 2)

Charge q2 is now displaced a distance y2 = 2.7 cm in the positive y-direction. What is the new value for the xcomponent of the force that q1 exerts on q2? -16.45

N

We use the Coulomb force equation here again to find the net force after using the Pythagorean theorem to obtain the distance between the points. Then we can use trig to find the angle and then the X component.

3)

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 5.71 N and points away from q1 and q3. What is the value (sign and magnitude) of the charge q3? 1.17

μC

Using the same steps as in problem (2) we can find the Coulomb force due to q1. This will be along the line from q1 to q2. Knowing the total force we can find the force from q3. Since Fq3 is positive then q3 must also be positive. Then by reversing the Coulomb equation, and knowing Fq3, we can find the charge value. 4) How would you change q1 (keeping q2 and q3 fixed) in order to make the net force on q2 equal to zero? Increase its magnitude and keep its sign the same As we saw above, adding together Fq1 and Fq3 gives a positive value so we need to increase the negative value i.e. q1. 5) How would you change q3 (keeping q1 and q2 fixed) in order to make the net force on q2 equal to zero? Decrease its magnitude and keep its sign the same Same argument as above but we can decrease the positive contribution by decreasing the positive q3 charge.

Point Charges in Two Dimensions

Three charges (q1 = 5.8 μC, q2 = -5.9 μC, and q3 = 3.4μC) are located at the vertices of an equilateral triangle with side d = 9.3 cm as shown. 1)

What is F3,x, the value of the x-component of the net force on q3? 20.7

N We can make use of the Coulomb equation to find the force from each charge combination.

The forces can then be added together componentwise. Θ here is 60°. Here the forces will add because attraction between q3 and q2 pulls it to the right while repulsive interaction with q1 pushes q3 to the right. 2) What is F3,y, the value of the y-component of the net force on q3? -0.306

N The same ideas as problem (1) but with the important difference that the Y components from q2 and q1 act in different directions and thus will subtract from one another. 3)

A charge q4 = 3.4 μC is now added as shown. What is F2,x, the x-component of the new net force on q2? -56.47

N We use the same principles here to add the x-components of the forces together. Note that we’re looking at the effect on q2 now and that q3 are identical and symmetric. 4) What is F2,y, the y-component of the new net force on q2? 0

N The forces from q2 and q3 cancel out due to symmetry about the x-axis. The force from q1 has no ycomponent and thus the total is zero. 5)

What is F1,x, the x-component of the new net force on q1? 15.08

N

Use the same methods as in (3) but note that q1 is positive and it’s position is different from q2 so while the forces will be the same, how they sum together will be different. 6) How would you change q1 (keeping q2,q3 and q4 fixed) in order to make the net force on q2 equal to zero? Decrease its magnitude and change its sign Since we saw in (3) the sign is negative we would need to either increase the force to the right from charges q3 and q2 or decrease the force in the negative direction by decreasing the charge of q 1.