High School Physics · Question of the Day

High School Physics Question of the Day

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Friday, November 6, 2026

A point charge Q=−3.0 μCQ=-3.0\,\mu\text{C} is fixed. Point PP is 0.50 m0.50\,\text{m} away from the charge. Using E=k∣Q∣/r2E = k|Q|/r^2 with k=9.0×109 N m2/C2k=9.0\times10^9\,\text{N}\,\text{m}^2/\text{C}^2, what is the magnitude of the electric field at PP?

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A point charge Q=−3.0 μCQ=-3.0\,\mu\text{C} is fixed. Point PP is 0.50 m0.50\,\text{m} away from the charge. Using E=k∣Q∣/r2E = k|Q|/r^2 with k=9.0×109 N m2/C2k=9.0\times10^9\,\text{N}\,\text{m}^2/\text{C}^2, what is the magnitude of the electric field at PP?

  1. 1.1×105 N/C1.1\times10^5\,\text{N/C} (correct answer)
  2. 1.1×104 N/C1.1\times10^4\,\text{N/C}
  3. 5.4×105 N/C5.4\times10^5\,\text{N/C}
  4. 5.4×104 N/C5.4\times10^4\,\text{N/C}

Explanation: This question tests understanding of modeling electric and magnetic fields and predicting how they interact with charges or currents. An electric field E exists in the region around a charge or group of charges and is defined as the force per unit charge that a positive test charge would experience at each point: E = F/q, measured in N/C (Newtons per Coulomb). The field direction is the direction a positive charge would be pushed (away from positive source charges, toward negative source charges), and field magnitude for a point charge is E = kQ/r², decreasing with the square of distance from the source charge Q. For a negative point charge Q = -3.0 μC, the electric field magnitude at distance r = 0.50 m is calculated using E = k|Q|/r² = (9×10⁹)(3×10−63×10^{-6})/(0.50)² = (9×10⁹)(3×10−63×10^{-6})/0.25 = (2.7×1042.7×10^4)/0.25 = 1.08×10^5 ≈ 1.1×10^5 N/C, and the direction is radially inward for negative charge; a positive test charge q placed at this point would experience force F = qE toward the charge, while a negative test charge would experience equal magnitude force in the opposite direction. Choice A is correct because it correctly calculates field strength using E = k|Q|/r² with proper values and units. Choice B claims a lower magnitude by possibly using r instead of r² in the denominator or miscalculating the exponent, when actually the inverse square law requires division by r², leading to a stronger field closer to the charge. Common mistakes to avoid: (a) assuming electric field points from negative to positive (it's opposite: from + to -), (b) forgetting that force on negative charge is opposite to field direction (F = qE with q < 0 reverses direction), (c) using left hand instead of right for right-hand rules (gives opposite result), (d) thinking magnetic field lines start and stop like electric ones (magnetic field lines are always closed loops), (e) assuming magnetic force is parallel to field or velocity (it's perpendicular to both), and (f) adding field vectors as scalars without considering direction (vector addition requires accounting for whether fields reinforce or cancel based on directions).