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Electric Potential Question

by blablasometin
Tags: electric potential
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blablasometin
#1
May5-10, 08:19 PM
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)
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#2
May5-10, 08:34 PM
P: 13
[removed]
blablasometin
#3
May5-10, 08:53 PM
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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