How Do You Calculate Electric Potential Energy with Given Charges and Distance?

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In summary: The potential energy of the system is the potential energy of the first charge plus the potential energy of the second charge.
  • #1
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.
 
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  • #2
bobsmith76 said:
V = (EQ)/q
V = kq/r
ΔV = W/q

Write down formula of potential energy and potential side by side,
then you'll know a much simpler method, and btw your calculation are fine, it doesn't mater how small a charge is

smallest possible free charge is e = 1.6 * 10-19
 
  • #3
bobsmith76 said:
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.

Make sure that you know the difference between the potential at a location and potential energy of the system as a whole. The problem statement suggests that the second charge is being brought to the location where the potential due to the first charge is 4.8V, 4.2 cm from the first charge; it's not being added to the first charge.
 
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