Q₁ = 6.0 nC is at (0.30 m, 0); Q₂ = -1.0 nC is at (0, 0.10 m); Q₃ = 5.0 nC is at (0, 0). What is the magnitude of the net force on the 5.0 nC charge?

Question

Quizplus · Accepted Answer

We can use Coulomb's Law to find the force between each pair of charges:

- Force between 5.0 nC and 6.0 nC: F1 = k * (5.0 nC) * (6.0 nC) / (0.30 m)^2 = 0.6 N repulsive (since they have the same sign).
- Force between 5.0 nC and -1.0 nC: F2 = k * (5.0 nC) * (1.0 nC) / (0.10 m)^2 = 22.5 N attractive (since they have opposite signs).
- Force between 6.0 nC and -1.0 nC: F3 = k * (6.0 nC) * (1.0 nC) / (0.30 m)^2 = 0.8 N attractive (since they have opposite signs).

The net force on the 5.0 nC charge is the vector sum of these three forces. Since F1 and F3 are in opposite directions and similar magnitudes, they cancel each other partially. The resulting force is given by:

Fnet = sqrt(F2^2 + (F1 - F3)^2) = sqrt(22.5^2 + (0.6 - 0.8)^2) = 5.4 × 10^-6 N

Therefore, the answer is C.

Q1 = 6

Q₁ = 6