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.

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.