Potential Difference and Kinetic Energy

In summary, the conversation discusses how to calculate the final kinetic energy of a helium nucleus moving through a potential difference. The equation E=qv is suggested, and it is noted that the nucleus will either gain or lose kinetic energy depending on the negative charge and potential difference. The conversation concludes with the solution of adding the initial and potential energies together to find the final kinetic energy.
  • #1
aimeemarie
2
0
Hello out there! I am having trouble figuring out how to set this up:

A helium nucleus (charge +2e) moves through a potential difference V = −0.70 kV. Its initial kinetic energy is 3.80 10-16 J. What is its final kinetic energy?

I feel like I should be using 1/2 mv^2=qV or E=qv. I feel like maybe I am close but I am getting lost. Can anyone help explain this please??
 
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  • #2
Welcome to PF!

Just use E=qv. You aren't asked for velocity. Remember E=qv will be added onto the initial energy.
 
  • #3
Be sure to keep track of your negative signs.
Given you have a negative charge moving through a net negative potential difference...visualize: your nucleus moves from 0.7V (+) to 0V (-). Is it going to slow down (lose kinetic energy) or speed up (gain energy, the field associated with that voltage doing work on it) as it moves through that difference?
 
  • #4
I can't believe I didn't think to add them together, thank you :)
 
  • #5


Sure, I'd be happy to help explain this problem to you. Let's start by breaking down the given information. We know that the helium nucleus has a charge of +2e, meaning it has a charge of 2 times the elementary charge, which is 1.6 x 10^-19 C. We also know that it is moving through a potential difference of -0.70 kV, or -700 V. Finally, we are given its initial kinetic energy of 3.80 x 10^-16 J.

Now, to solve for its final kinetic energy, we can use the equation you mentioned, E=qV. In this case, q represents the charge of the particle and V represents the potential difference it is moving through. So, plugging in the values we know, we get:

E = (2e)(-700 V)
E = -1.4 x 10^-16 J

This is the change in kinetic energy, but we want to find the final kinetic energy. To do this, we need to add the initial kinetic energy to the change in kinetic energy. So, the final kinetic energy would be:

3.80 x 10^-16 J + (-1.4 x 10^-16 J) = 2.4 x 10^-16 J

Therefore, the final kinetic energy of the helium nucleus is 2.4 x 10^-16 J. I hope this helps clarify the problem for you. Remember, when working with potential difference and kinetic energy, it's important to pay attention to the signs and units of the values given. Good luck!
 

1. What is potential difference?

Potential difference, also known as voltage, is the difference in electrical potential between two points in a circuit. It is measured in volts (V) and is the driving force that causes electric charges to flow.

2. How is potential difference related to kinetic energy?

Potential difference is directly related to kinetic energy through the movement of electric charges. As charges move through a circuit, they gain kinetic energy from the potential difference and are able to do work, such as powering a light bulb or motor.

3. What is the formula for calculating potential difference?

The formula for potential difference is V = W/Q, where V is potential difference, W is the work done on the charge, and Q is the amount of charge. This can also be represented as V = IR, where I is the current flowing through the circuit and R is the resistance of the circuit.

4. How does potential difference affect the flow of electric current?

The greater the potential difference, the greater the force pushing electric charges to flow through a circuit. As potential difference increases, the rate of flow of electric charge, or current, also increases. Conversely, a lower potential difference will result in a slower rate of current flow.

5. Can potential difference be negative?

Yes, potential difference can be negative. This occurs when the direction of the electric field is opposite to the direction of the electric current. In this case, the charges are moving from a higher potential to a lower potential, resulting in a negative potential difference.

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