Question Detail
Question
What is the magnitude of a point charge whose electric field 50 cm away has magnitude 2.0 N/C?
Correct Answer
Option C is the correct answer.
Detailed Explanation
The magnitude of the electric field (E) due to a point charge (q) at a distance (r) is given by Coulomb's law for electric fields:E = k * |q| / r2Where k is Coulomb's constant (approximately 9 x 109 N·m²/C²).Given E = 2.0 N/C and r = 50 cm = 0.50 m.Rearranging the formula to solve for |q|:|q| = E * r2 / k|q| = (2.0 N/C) * (0.50 m)2 / (9 x 109 N·m²/C²)|q| = (2.0) * (0.25) / (9 x 109) = 0.5 / (9 x 109) = 0.0555... x 10-9 C = 5.55... x 10-11 C.Rounding to two decimal places, |q| ≈ 5.56 x 10-11 C.
Hint
Use Coulomb's law for the electric field of a point charge: E = k|q|/r2. Remember to convert distance to meters.