Two point charges of +1 nC are located at (0, +5 cm) and (0, -5 cm). What is the electric field at the midpoint (0, 0)?

• A. 0 N/C
• B. 2510 N/C upward
• C. 3596 N/C downward
• D. 7192 N/C upward

Answer: (A)

The key idea is vector superposition: fields from multiple charges add up, taking direction into account. The two +1 nC charges sit symmetrically above and below the origin, each 0.05 m away. At the midpoint the distance to either charge is the same, so the magnitudes of their fields there are equal.

For a positive charge, the field points away from the charge. The charge at +0.05 m produces a field at the origin that points downward, while the charge at −0.05 m produces a field that points upward. Since the magnitudes are identical, these two fields cancel.

Quantitatively, each field has magnitude E = k q / r^2 = (8.99×10^9 N·m^2/C^2)(1×10^-9 C) / (0.05 m)^2 ≈ 3.60×10^3 N/C. One is downward, the other upward, so the net electric field at the midpoint is 0 N/C.