Two fixed charges +Q at x = -d/2 and -Q at x = +d/2. A test charge q is placed at the midpoint (x = 0). What is the electric force on the test charge?

• A. 0 N
• B. Nonzero depending on q
• C. It depends on distance only
• D. It is always attracted to the positive charge

Answer: (A)

The test charge feels the vector sum of the Coulomb forces from both fixed charges. At the midpoint, each fixed charge is a distance d/2 away. The magnitude of each individual force is F0 = k |q| Q /(d/2)^2 = 4k |q| Q / d^2. The left charge is +Q, which would push the test charge to the right if q is positive, while the right charge is -Q, which attracts the test charge toward itself; from the origin, that attraction also lies to the right. So the two forces add along the same direction, giving a total force F = 2F0 = 8k |q| Q / d^2 to the right for q > 0 (and to the left for q < 0). In any case, the net force is nonzero as long as q ≠ 0 and it depends on q (and on Q and d through the 1/d^2 dependence). It is not determined by distance alone, and it does not imply a universal attraction to the positive charge.