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bobsmith76
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Homework Statement
The electric potential difference due to a point charge is 4.8 V at a distance of 4.2 cm from the charge. What will be the electric potential energy of the system if a second charge of +6.0 μC is placed at that location?
Homework Equations
V = (EQ)/q
V = kq/r
ΔV = W/q
The Attempt at a Solution
I've tried all three equations but none of them give me the right answer.
I try to find the charge of the first point
step 1. 4.8 = kx/.042
step 2. .042(4.8) = kx
step 3 (.042(4.8))/k = x
step 4 = 2.24 * 10^-11 (that seems like a ridiculously small number, most charges are 10^-6. Adding 6*10^-6 will do nothing to the number, so I must be doing something wrong. I just can't figure out what. Even if I do get the right number and I add the charge I can't figure out what equation to use next.