Question

If the potential due to a point charge is 5.20 ✕ 10² V at a distance of 14.7 m, what are the sign and magnitude of the charge? (Enter your answer in C.)

 C

Answer 1

Electric potential is a scalar quantity which is given by:

V = k*Q/R

Since it's scalar quantity, So if potential due to point charge is positive, then charge will be positive and if negative than charge will be negative.

Now here given that Potential, V = 5.20*10^2 V, So

Charge will have positive sign.

At a distance R = 14.7 m, we know that:

V = k*Q/R

Q = V*R/k

k = electrostatic constant = 9*10^9

So using above values:

Q = 5.20*10^2*14.7/(9*10^9)

Q = 8.49*10^-7 C = magnitude of point charge

Let me know if you've any query.