Asked by Brayden
A point charge has an excess of 5.0 × 1012 electrons. What would be the electric potential at a distance of 0.50 m from the charge? (e = 1.6 × 10-19 C)
Answers
Answered by
Anonymous
It is in your physics text !
Answered by
Brayden
wdym
Answered by
Brayden
nvm thanks
Answered by
Anonymous
can you help me out I don't know what you mean in the text
Answered by
Dan
Givens: N = 5.0 x 10^12, r = 0.50 m, e = 1.6 x 10^-19
First use the equation q = Ne to find the charge.
q = (5.0 x 10^12) x (-1.6 x 10^-19)
q = -8.0 x 10^-7
The elementary charge here is “negative” because there is an excess of electrons. If the object had lost electrons, the balloon would have a deficit of electrons, e.g., more protons than electrons. The elementary charge would then be “positive”.
Then use the equation V = kq/r to find electric potential
V = (9 x 10^9) x (-8.0 x 10^-7) / 0.50 m
V = -1.44 x 10^4 V
First use the equation q = Ne to find the charge.
q = (5.0 x 10^12) x (-1.6 x 10^-19)
q = -8.0 x 10^-7
The elementary charge here is “negative” because there is an excess of electrons. If the object had lost electrons, the balloon would have a deficit of electrons, e.g., more protons than electrons. The elementary charge would then be “positive”.
Then use the equation V = kq/r to find electric potential
V = (9 x 10^9) x (-8.0 x 10^-7) / 0.50 m
V = -1.44 x 10^4 V
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