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PHY 102 Test Compilation PHY 102 12 Objective Question

Question

What is the magnitude of a point charge whose electric field 50 cm away has magnitude 2.0 N/C?
Options
A 3.76 x 10<sup>-11</sup> C
B 4.05 x 10<sup>-11</sup> C
C 5.56 x 10<sup>-11</sup> C
Correct Answer
D 2.00 x 10<sup>-11</sup> 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.

Question Info