Question

Two point charges are located on the x-axis. One has a charge of 1.77 μC and is located at x = 0.0 m, and the other has a charge of -4.09 μC and is located at x = 15.1 m. At what location on the x-axis (other than at infinity) would the electric force on a third point charge of 3.32 μC be zero?

Answer 1

Answer:

x = -29.032m

Explanation:

Since the third charge is positive, it cannot be between the other two charges, because it would be repelled by the positive one and attracted by the negative one, so the electric force would never be zero.

This leaves only two options: To the left of the positive one or to the right of the negative one.

If it was located on the right of the negative charge, the force of the positive charge would be weaker because of both the distance is larger and its charge is smaller than the negative charge. So, there is no point the would make the result force equal zero.

This means that the third charge has to be at the left of the positive charge. With this in mind, we make the calculations:

$F_{13}=K*\frac{Q_{1}*Q_{3}}{d^{2}} =F_{23}=K*\frac{Q_{2}*Q_{3}}{(x_{2}+d)^{2}}$

Replacing the values of Q1=1.77, Q2=4.09, X2=15.1, we solve for d and get two possible results:

d1 = 29.032m   and d2 = -5.99m

Since we assumed in our formula that the third charge was on the left of the positive charge, the distance d has to be positive so that our final result can be a negative position. This is X = -d

This way, we get:

X = -29.032m