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Suppose the potential due to a point charge is 6.25x10^2 v at a distance of 17m. What is the magnitude of the charge, in coulombs?

User Itchydon
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1 Answer

4 votes

Answer:


q=1.18*10^(-6)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 solve to find q:


q=(V*r)/(k)=(6.25*10^(2)*17)/(8.99*10^(9))=1.18*10^(-6)C

User Shourav
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