Three charges form an equilateral triangle with side length 1 cm. q1 = 1 nC at one corner, q2 = 2 nC at the next corner, and q3 = 2 nC at the third corner. What is the magnitude of the force on q1 due to the other two charges?

• A. 1.56×10^-4 N
• B. 2.88×10^-4 N
• C. 3.12×10^-4 N
• D. 6.24×10^-4 N

Answer: (C)

Electric forces from multiple charges add as vectors, so the force on q1 is the vector sum of the forces due to q2 and q3.

Compute each force: F from q1 to q2 is k times q1 q2 over r^2. With q1 = 1 nC = 1×10^-9 C, q2 = 2 nC = 2×10^-9 C, r = 1 cm = 0.01 m, k ≈ 9.0×10^9 N m^2/C^2,

F12 = k q1 q2 / r^2 = 9.0×10^9 × (1×10^-9)(2×10^-9) / (1×10^-4) = 1.8×10^-4 N.

The same magnitude comes from q3 since q3 = 2 nC and is located at the other corner.

The two forces on q1 are along lines to q2 and to q3, separated by 60°, since the triangle is equilateral. The resultant magnitude uses vector addition: F_total^2 = F12^2 + F13^2 + 2 F12 F13 cos(60°). With F12 = F13 = 1.8×10^-4 N and cos(60°) = 1/2, this gives F_total^2 = 2(F^2) + F^2 = 3F^2, so F_total = F × sqrt(3).

Numerically, F_total ≈ 1.8×10^-4 N × 1.732 ≈ 3.12×10^-4 N.

So the force on q1 due to the other two charges is about 3.12×10^-4 N, directed along the angle bisector away from both charges.