Electric Potential Energy Problem

In summary: From there, calculate the electric field at the origin. Then calculate the electric potential energy. Compare the two values and see if you get the same answer.
  • #1
KayDe
2
0

Homework Statement



Two Charged Objects (Q1= -9.0nC & Q2= 15nC) are fixed in place on the +x-axis. The first charge is 12 cm to the left from the origin the other is 12 cm to the right from the origin. A proton is moved from the origin to a position on the +y-axis 9.0 cm from the origin. a) What is the change in potential energy of the proton due to Q1 & Q2? [/B]

Homework Equations


As given in our Course

F=k |(q1q2)| / r2
E=k |q| / r
Potential Energy (uniform electric field) = qEΔd
Potential Energy (not in uniform electric field) = KQq(1/rf-1/ri)
[rf= final distance ; ri= initial distance]

The Attempt at a Solution


[/B]
I tried calculating the electric field at the origin between the two points due to the two charged objects. I figured I needed to calculate the electric field since there is not a uniform electric field with two different point charges present. With the electric field I then tried to calculate the electric potential energy since I know the distance and the charge of the proton. I then did the same for the final position and then calculated the change in the potential energy. The given solution for this problem is 1.44 x 10-17 J which I did not get. I either am approaching this the wrong way, understanding it wrong or calculating the wrong things but I get a different answer every time and I'm not sure what I'm missing :cry:. Thank You
 
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  • #2
You won't need the values of the electric field here. You just need the electric potential energy of the proton in its initial and final positions.
 
  • #3
gneill said:
You won't need the values of the electric field here. You just need the electric potential energy of the proton in its initial and final positions.
I tried that too but I'm still not getting the right solution :/
 
  • #4
Can you show your work in detail? Start with the potential energy at the origin.
 

1. What is electric potential energy?

Electric potential energy is the energy a charged particle possesses due to its position in an electric field. It is the potential energy that can be converted into kinetic energy as the particle moves in the direction of the electric field.

2. How is electric potential energy calculated?

The electric potential energy of a charged particle is calculated by multiplying its charge (q) with the electric potential (V) at its position in the electric field. The equation is U = qV, where U is the electric potential energy in joules (J), q is the charge in coulombs (C), and V is the electric potential in volts (V).

3. What factors affect the electric potential energy of a charged particle?

The electric potential energy of a charged particle is affected by its charge, the electric potential at its position, and the distance between the particle and the source of the electric field. As the charge or electric potential increases, the potential energy also increases. However, as the distance between the particle and the source increases, the potential energy decreases.

4. Can electric potential energy be negative?

Yes, electric potential energy can be negative. This occurs when the charged particle and the source of the electric field have opposite charges and are attracted to each other. In this case, the electric potential energy is negative because work is done by the electric field on the particle as it moves towards the source.

5. How is electric potential energy related to electric potential difference?

Electric potential energy and electric potential difference are related through the equation U = q∆V, where U is the electric potential energy, q is the charge, and ∆V is the change in electric potential. This means that the change in electric potential energy of a particle is directly proportional to the change in electric potential difference between its initial and final positions.

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