Four charges are arranged with q1 = 1 nC at the bottom, q2 = 2 nC, q3 = -6 nC, q4 = 2 nC with q2 above-left of q1, q4 above-right of q1, and q3 above q1 between q2 and q4 (the angle from q2 to q4 is 90 degrees). What is the magnitude of the net force on the bottom charge q1 and its direction?

• A. 1.14×10^-5 N, 90 degrees
• B. 1.01×10^-5 N, 0 degrees
• C. 2.28×10^-5 N, 90 degrees
• D. 3.00×10^-5 N, 45 degrees

Answer: (A)

The forces on q1 come from the other three charges and must be added as vectors (Coulomb’s law). Because q2 and q4 lie symmetrically about the vertical through q1, their horizontal components cancel. The vertical components of their repulsive forces point downward. The negative charge q3 sits above q1 and attracts q1 upward toward itself, giving a vertical force upward.

Let the distance from q1 to q3 be a. Then q2 and q4 are at a√2 away from q1, since they form 45° directions from the vertical. The force magnitudes are:

• F2 = F4 = k q1 q2 / (a√2)^2 = k q1 q2 /(2 a^2), with vertical components downward of F2_y = F2 · (a / (a√2)) = F2 / √2 for each, so combined downward from q2 and q4 is √2 F2 = k q1 q2 /(√2 a^2).

• F3 = k q1 |q3| / a^2, directed upward toward q3.

The net force on q1 is purely vertical and given by

F_y = F3 − (downward from q2 and q4) = k q1 |q3| / a^2 − k q1 q2 /(√2 a^2)

= (k q1 / a^2) [|q3| − q2/√2].

Plugging in q1 = 1 nC, q2 = 2 nC, |q3| = 6 nC, and using the distance a read from the diagram, the value comes out to about 1.14 × 10^-5 N directed upward (90° from the +x axis).