Potential Difference and Kinetic Energy

[aimeemarie]

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??

 

[Delphi51]

Welcome to PF!

Just use E=qv. You aren't asked for velocity. Remember E=qv will be added onto the initial energy.

 

[Apphysicist]

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?

 

[aimeemarie]

I can't believe I didn't think to add them together, thank you :)

 

[rwisz]

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!