If the potential due to a point charge is [tex]6.00 \times 10^2 \, \text{V}[/tex] at a distance of [tex]14.9 \, \text{m}[/tex], what are the sign and magnitude of the charge? (Enter your answer in C.)

A) [tex]-4.03 \times 10^{-8} \, \text{C}[/tex]
B) [tex]4.03 \times 10^{-8} \, \text{C}[/tex]
C) [tex]-6.00 \times 10^{-8} \, \text{C}[/tex]
D) [tex]6.00 \times 10^{-8} \, \text{C}[/tex]

Option b, the magnitude of the charge is 4.03 × 10^-8 C and its sign is positive. This is found by using the formula for electric potential due to a point charge, and the positive potential indicates a positive charge.

Explanation:

If the potential due to a point charge is 6.00 × 102 V at a distance of 14.9 m, we can use the formula for electric potential (V) due to a point charge (Q) at a distance (r): V = kQ/r, where k is the Coulomb's constant (8.99 × 109 Nm2/C2). To find the magnitude of the charge, we can rearrange the formula to solve for Q: Q = Vr/k.

By substituting the given values into the formula, we get Q = (6.00 × 102 V × 14.9 m) / (8.99 × 109 Nm2/C2). This calculation results in Q = 4.03 × 10-8 C. Since the potential is positive, the charge is also positive.

Therefore, the magnitude of the charge is 4.03 × 10-8 C and its sign is positive.