Potential Difference and Kinetic Energy

AI Thread Summary
To find the final kinetic energy of a helium nucleus moving through a potential difference of -0.70 kV, use the equation E = qV, where q is the charge and V is the potential difference. The initial kinetic energy is 3.80 x 10^-16 J, and since the nucleus has a charge of +2e, the energy change due to the potential difference must be added to the initial kinetic energy. As the nucleus moves through a negative potential difference, it gains kinetic energy. Therefore, the final kinetic energy is the sum of the initial kinetic energy and the energy change from the potential difference.
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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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Just use E=qv. You aren't asked for velocity. Remember E=qv will be added onto the initial energy.
 
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?
 
I can't believe I didn't think to add them together, thank you :)
 
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