Electric Potential Energy and Potential Difference
Solution:
1. a) Given that the potential difference between points a and b is given by DV_{ }= 400 V, we can then write
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b) To increase the potential energy between two charges of the same polarity, move them towards each other. Therefore, move the charge in direction 4 (choice D).
2. a) To determine the speed of the proton we must first determine its change in potential energy and hence its change in kinetic energy. Notice that the question states that the electric potential difference was 12000 V. Provincial questions generally state the absolute value of the electric potential difference so be careful.
DE_{P} = qDV = DE_{K }
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b) In this question you are to assume that at the point where proton exits the parallel plates, the electric potential energy due to the 5.0mC charge is 0 J. Notice the diagram says “not to scale”, and gives no indication as to how far the 5.0mC charge is from the parallel plates, so we set E_{Pi} = 0 J. We also know that E_{Kf} = 0 J, since the proton is brought to a stop. Thus we have a conservation of energy problem to solve, where E_{Ki } = E_{Pf}.
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