SkaterGirl6398 SkaterGirl6398
  • 14-09-2019
  • Physics
contestada

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?

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jorgezarate
jorgezarate jorgezarate
  • 17-09-2019

Answer:

[tex]q=+8.34*10^{-7}C}[/tex]

Explanation:

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

[tex]V=k\frac{q}{r}[/tex]

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

We replace the values in order to find q:

[tex]q=\frac{V*r}{k}=\frac{500*15}{8.99*10^{9}}=8.34*10^{-7}C[/tex]

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Hihihihihihihihiihih
Hihihihihihihihiihih Hihihihihihihihiihih
  • 21-12-2021

Answer:

i apolagize im late but yeah bois 700 points

Explanation:

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