If the potential due to a point charge is 5.00 x 10^2 V at a distance of 15.0 m, what are the sign and magnitude of the charge?

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?

Question by OpenStax is licensed under CC BY 4.0.

Final Answer

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

Please note: the video incorrectly shows the wrong exponent for 10. It should be $-7$, not $-6$.

Solution Video

OpenStax College Physics Solution, Chapter 19, Problem 29 (Problems & Exercises) (0:37)

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Calculator Screenshots

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Video Transcript

This is College Physics Answers with Shaun Dychko. The potential from a point charge is Coulombs 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 then, we get the charge, it's R V over K. So, that's 15 meters times 500 volts divided by Coulomb's constant, 8.99 times ten to the nine Newton meters squared per Coulomb squared. And, that gives a charge of 8.34 times ten to the minus six Coulombs. This is a positive charge because the potential is positive 500 volts.