Two charges q1 = -20.0 nC at x1 = -1.685 m and q2 = 34.0 nC at x2 = 0 m. A third charge q3 = 46.5 nC is placed at x3 = -1.200 m. What is the x-component of the net force on q3?

• A. +4.54e-5 N
• B. -4.54e-5 N
• C. -1.00e-4 N
• D. 0 N

Answer: (B)

The x-component of the net force on q3 comes from the two Coulomb forces due to q1 and q2, and since all charges lie on the x-axis, the forces lie along x as well.

Compute distances along the axis:

• r31 = x3 − x1 = −1.200 − (−1.685) = 0.485 m

• r32 = |x3 − x2| = |−1.200 − 0| = 1.200 m

Use Coulomb’s law, F = k q3 q? / r^2, with direction given by the sign of the interaction.

• Force on q3 due to q1: q1 is negative and q3 is positive, so this is attractive toward q1 (to the left). Magnitude:

F31 = k |q3 q1| / r31^2

= (8.99×10^9) × (46.5e−9 × 20e−9) / (0.485)^2

≈ 3.56×10^−5 N

Direction: negative x.

• Force on q3 due to q2: q2 is positive and q3 is positive, so this is repulsive away from q2 (toward the left since q2 is to the right). Magnitude:

F32 = k |q3 q2| / r32^2

= (8.99×10^9) × (46.5e−9 × 34e−9) / (1.200)^2

≈ 9.87×10^−6 N

Direction: negative x.

Add the x-components:

F_x ≈ −3.56×10^−5 − 9.87×10^−6 ≈ −4.55×10^−5 N

So the net force on q3 points to the left with magnitude about 4.54×10^−5 N, i.e., −4.54×10^−5 N along the x-axis.