Three point charges, +6.0 µC, +1.7 µC, and

Question

−2.3 µC, lie along the x-axis at 0 cm, 1.7 cm,
and 6.0 cm, respectively.
What is the force exerted on q1 by the other
two charges? (To the right is positive.) The
Coulomb constant is 8.99 × 10
9
N · m2
/C
2
.
Answer in units of N

Answer 1

First, let's find the force exerted by q2 on q1:

F21 = (k * q1 * q2) / r21^2

where k is the Coulomb constant, q1 and q2 are the charge magnitudes, and r21 is the distance between them.

F21 = (8.99 * 10^9 N * m^2 / C^2) * (+6.0 * 10^-6 C) * (+1.7 * 10^-6 C) / (1.7 * 10^-2 m)^2
F21 = 3.69 N (to the right)

Now, let's find the force exerted by q3 on q1:

F31 = (k * q1 * q3) / r31^2

where r31 is the distance between q1 and q3.

F31 = (8.99 * 10^9 N * m^2 / C^2) * (+6.0 * 10^-6 C) * (-2.3 * 10^-6 C) / (6.0 * 10^-2 m)^2
F31 = -1.15 N (to the left)

The net force on q1 is the sum of these two forces:

F_net = F21 + F31
F_net = 3.69 N + (-1.15 N) = 2.54 N (to the right)

So the force exerted on q1 by the other two charges is 2.54 N (to the right).