What is the electric potential at a distance of 0.30 m from a point charge of 6.0 nC (k = 9.0 × 10⁹ N m² C⁻²)?
Correct answer B. 180 V
Explanation
V = kq/r = 9.0 × 10⁹ × 6.0 × 10⁻⁹ / 0.30 = 54/0.30 = 180 V. Dividing by r² as for field strength gives 600 V.
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