# Electric Potential Question

by blablasometin
Tags: electric potential
 P: 2 A proton's speed as it passes point A is 53,000 m/s. It follows the trajectory shown in the figure below, in which V1 = 15 V and V3 = 5 V. What is the proton's speed at point B? Figure: http://www.webassign.net/knight/p29-44alt.gif 1. The problem statement, all variables and given/known data Mass of proton: 1.67e-27 Charge of proton: 1.6e-19 Difference in Electric Potential velocity at point A: 53000m/s V1 = 15V V2 = 10V V3 = 5V 2. Relevant equations KE=1/2mv^2 V=(kq)/r V=Ed 3. The attempt at a solution 10q=1/2mv^2 v=sqrt(20*1.6e-19/1.67e-27)
 P: 13 [removed]
 P: 2 never mind I got the solution: KE(B) + PE(B) = KE(A) + PE(A) (1/2)mvB^2 + qV(B) = (1/2)mvA^2 + qV(A)\ vB^2 = vA^2 + (2q/m)[V(A) - V(B)] Answer: 68739.85m/s I should have known it was a simple conservation of energy problem

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