If the potential due to a point charge is 5.15 ✕ 102 V at a distance of 14.8 m, what are the sign and magnitude of the charge? (Enter your answer in C.)

q = 8.47×10⁻⁷ C.

Explanation:

From Coulomb's,

V = kq/r Equation 1

Where, V = Electric potential, q = charge, r = distance, k = proportionality constant.

Make q the subject of the equation

q = Vr/k Equation 2

Given: V = 5.15×10² V, r = 14.8 m

Constant: k = 9.0×10⁹ Nm²/C²

Substitute into equation 2

q = 5.15×10²×14.8/9.0×10⁹

q = 8.47×10⁻⁷ C.

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