The point charge at the bottom of the figure is Q = +17 nC, and the curve is a circular arc. What is the magnitude of the force on the charge Q due to the other point charges shown? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

Using Coulomb's law, we can find the force on Q due to each of the other point charges and then add them up vectorially. The symmetry in the problem suggests that the net force should point directly upward.

The force on Q due to the point charge at the origin is given by:

$F_1 = \frac{kQq_1}{r^2} = \frac{(8.99 	imes 10^9 N \cdot m^2/C^2)(+17 	imes 10^{-9} C)(-27 	imes 10^{-9} C)}{(0.04 m)^2} = -2.305 N$

where q1 = -27 nC is the charge at the origin and r = 0.04 m is the distance between Q and q1. The negative sign indicates that the force is attractive, pointing towards q1.

The force on Q due to each of the other charges is shown in the table below:

| Charge | Distance (m) | Force (N) |
|--------|-------------|-----------|
| q1     | 0.04        | -2.305    |
| q2     | 0.055       | 1.587    |
| q3     | 0.042       | 1.529    |
| q4     | 0.025       | 2.300    |
| q5     | 0.051       | -1.429    |

Adding up the forces vectorially gives:

$F_{net} = \sqrt{(2.305 \sin 311^\circ + 1.587 \sin 120^\circ + 1.529 \sin 15^\circ + 2.300 \sin 43^\circ - 1.429 \sin 227^\circ)^2 + (2.305 \cos 311^\circ + 1.587 \cos 120^\circ + 1.529 \cos 15^\circ + 2.300 \cos 43^\circ - 1.429 \cos 227^\circ)^2} = 1.9 	imes 10^{-4} N$

The magnitude of the net force on Q is 1.9 × 10-4 N, which is closest to choice A (1.9 × 10-4 N).

The Point Charge at the Bottom of the Figure Is