Electric Potential Energy & Speed of Charge: Q=+85.5 µC, q=-2.73 µC

In summary, the electric potential energy of the system of charges is -6.496597059 J. To calculate the speed of the second charge when it reaches (0.125 m, 0), you need to find the change in potential energy between the initial separation of 0.323m and the final separation of 0.125m. This will give you a final potential energy, and then you can use the formula v = \sqrt{2U/m} to find the speed.
  • #1
wr1015
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0
A point charge Q = +85.5 µC is held fixed at the origin. A second point charge, with mass m = 0.0526 kg and charge q = -2.73 µC, is placed at the location (0.323 m, 0).

(a) Find the electric potential energy of this system of charges.

(b) If the second charge is released from rest, what is its speed when it
reaches the point (0.125 m, 0)?


ok so i found part (a) -6.496597059 J but how do i calculate (b)?? I've tried subtracting the electric potential energy of the distance traveled from .323 to .125 (.198 m) from the electric potential energy of the system to find the change in electric potential energy. Then i divided that by my second charge (-2.73E-6) to get my change in potential so i could plug that into [tex]v= \sqrt{2q\Delta V/m}[/tex] and am getting wrong answers... any guess on where I'm going wrong??
 
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  • #2
wr1015 said:
A point charge Q = +85.5 µC is held fixed at the origin. A second point charge, with mass m = 0.0526 kg and charge q = -2.73 µC, is placed at the location (0.323 m, 0).

(a) Find the electric potential energy of this system of charges.

(b) If the second charge is released from rest, what is its speed when it
reaches the point (0.125 m, 0)?


ok so i found part (a) -6.496597059 J but how do i calculate (b)?? I've tried subtracting the electric potential energy of the distance traveled from .323 to .125 (.198 m) from the electric potential energy of the system to find the change in electric potential energy. Then i divided that by my second charge (-2.73E-6) to get my change in potential so i could plug that into [tex]v= \sqrt{2q\Delta V/m}[/tex] and am getting wrong answers... any guess on where I'm going wrong??
[itex]v = \sqrt{2U/m}[/itex] where U is the potential energy change.
I get the change in potential energy to be 10.3 J which gives a speed of 19.8 m/sec.

AM
 
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  • #3
wr1015 said:
(b)?? I've tried subtracting the electric potential energy of the distance traveled from .323 to .125 (.198 m) from the electric potential energy of the system to find the change in electric potential energy.

Initially, the charges are separated by 0.323m and have the potential energy that you calculated in part (a). Finally, the charges are separated by 0.125m and have potential energy = ?. You need to calculate the final potential energy, and then the difference between the initial and final potential energies.
 
  • #4
thank you all for your help
 

1. What is electric potential energy?

Electric potential energy is the potential energy that a charged object has due to its position in an electric field. It is measured in joules (J).

2. How is electric potential energy calculated?

Electric potential energy (U) is calculated by multiplying the charge (q) in coulombs (C) by the potential difference (V) in volts (V). The formula is U = qV.

3. What is the speed of charge?

The speed of charge refers to the velocity at which a charged object moves. It is measured in meters per second (m/s).

4. How do you calculate the speed of charge?

The speed of charge can be calculated by dividing the distance traveled (d) by the time it took to travel that distance (t). The formula is v = d/t.

5. Can you provide an example of calculating electric potential energy and speed of charge?

Using the given values of Q=+85.5 µC and q=-2.73 µC, we can calculate the electric potential energy by multiplying Q by the potential difference. If the potential difference is 10 V, then the electric potential energy would be U = (+85.5 µC)(10 V) = +855 µJ. To calculate the speed of charge, we would need to know the distance traveled and the time it took. Let's say the distance was 5 meters and it took 2 seconds to travel that distance. Then, the speed of charge would be v = (5 m)/(2 s) = 2.5 m/s.

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