Question

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

Final Answer

$+8.34 \times 10^{-7} \textrm{ C}$

### Solution video

# OpenStax College Physics, Chapter 19, Problem 29 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. At a distance 15.0 meters from a point charge, the potential is 500 volts and the question asks us to find out what is the sign and magnitude of the charge? So the formula for the potential due to a point charge is Coulomb's constant times the charge divided by the distance from it and we can solve this for Q by multiplying both sides by r over k and switching the sides around, we get Q equals rV over k and that's—a distance of—15.0 meters times —a potential of—500 volts divided by—Coulomb's constant of— 8.99 times 10 to the 9 Newton meters squared per Coulombs squared and that gives 8.34 times 10 to the minus 7 Coulombs and this charge is positive because it is a positive potential.