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Jabababa
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Homework Statement
What is the electric potential energy of a proton located 20.0 A (one angstrom or 1A is equal to 10^-10m) from another proton?
Homework Equations
Ep= F x d
F= kq1q2/r^2
The Attempt at a Solution
Ep= Fxd
F=kq1q2/r^2
therefore Ep = (Kq1q2)(d)/r^2
d=r
so Ep= KQ1Q2/r
(9x10^9)(1.60x10^-19)(1.60x10^-19)/(20x10^-10)
= 1.152x10^-19J
Help me check see if i got it right please!