Question

If the potential due to a point charge is (5.00 X 10^2 {V}) at a distance of (15.0 {m}), what are the sign and magnitude of the charge?

a) Positive, (1.11 X 10^{-9} {C})
b) Positive, (3.33 X 10^{-9} {C})
c) Negative, (1.11 X 10^{-9} {C})
d) Negative, (3.33 X 10^{-9} {C})

Answer 1

The sign and magnitude of the charge are positive and 1.11 × 10^(-9) C respectively.

Step-by-step explanation:

To determine the sign and magnitude of the charge, we can use the formula for potential due to a point charge: V = kQ/r, where V is the potential, k is the Coulomb's constant (9.0 × 10^9 Nm²/C²), Q is the charge, and r is the distance. Rearranging the formula to solve for Q, we have Q = Vr/k. Plugging in the given values, we get Q = (5.00 × 10² V)(15.0 m)/(9.0 × 10^9 Nm²/C²) = 1.11 × 10^(-9) C. Since the potential is positive, the charge must also be positive. Therefore, the answer is a) Positive, (1.11 × 10^(-9) C).