In a line of three charges with q_middle = -2 nC, q1 is 10 cm from the middle, and q2 is 10 cm from q1. If the force on q2 is zero, what must q1 be?

• A. -8 nC
• B. 8 nC
• C. 2 nC
• D. -2 nC

Answer: (B)

For the net force on q2 to be zero, the forces on q2 from q1 and from q_middle must cancel in magnitude and point in opposite directions.

Distance details: q1 is 10 cm from the middle, and q2 is 10 cm from q1, so the distance between q_middle and q2 is 20 cm. The force on q2 due to q1 has magnitude F1 = k |q1||q2| / (0.10 m)^2. The force on q2 due to q_middle has magnitude Fm = k |q_middle||q2| / (0.20 m)^2. Setting these equal in magnitude (and noting the directions must be opposite) gives:

|q1| / (0.10)^2 = |q_middle| / (0.20)^2

|q1| = |q_middle| (0.10^2 / 0.20^2) = (2 nC) (0.01 / 0.04) = 0.5 nC.

The sign must be such that the forces oppose; with q_middle = -2 nC, q1 should be positive (to create a force on q2 that can oppose the attraction toward the middle). Therefore q1 must be +0.5 nC.

Note: 0.5 nC is not among the listed options, so the given answer choice set appears inconsistent with this setup.