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

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To calculate electric potential energy when a second charge is placed in the electric field of a point charge, first understand that the potential difference (4.8 V) is due to the first charge at a distance of 4.2 cm. The potential energy (U) of the system can be calculated using the formula U = qV, where q is the second charge (+6.0 μC) and V is the potential at that location. It is important to differentiate between electric potential and potential energy, as the potential energy is the energy of the system when the second charge is introduced to the existing potential. The calculations provided indicate that the initial charge's value is correct, despite appearing small, and the second charge's impact on the potential energy should be calculated based on the established potential. Understanding these concepts will clarify the approach to solving the problem effectively.
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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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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
 
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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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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