Question

If the potential due to a point charge is 500 V at a distance of 15.0 m, what are the sign and magnitude of the charge?

Answer 1

i apolagize im late but yeah bois 700 points

Step-by-step explanation:

Answer 2

`q=+8.34*10^(-7)C}`

Step-by-step explanation:

The potential V due to a charge q, at a distance r, is:

`V=k(q)/(r)`

k=8.99×109 N·m^2/C^2 :Coulomb constant

We replace the values in order to find q:

`q=(V*r)/(k)=(500*15)/(8.99*10^(9))=8.34*10^(-7)C`