An electron moving parallel to the x axis has an initial speed of 3.70*10^6m/s a

Question

An electron moving parallel to the x axis has an initial speed of 3.70*10^6m/s at the origin. Its speed is reduced to 1.40*10^5m/s at the point x=2cm. A. Calculate the electric potential difference between the origin and that point. B. Which point has higher potential energy? C. Which point does the electron have a higher potential energy? D. If instead a proton had been traveling at the origin with the same velocity as the electron, how fast would it be moving when it reaches the point x=2cm. E. Which point has a higher potential? F. Which point does the proton have a higher potential energy?

Explanation / Answer

E (lost) = 1/2 m * (v1^2 - v2^2)  = V * q
Note that E(lost) is a positive quantity and q is negative so
V must also be negative (higher potential to lower potential)
V = 1/2 m * (v1^2 - v2^2) / q =
(9.11 * 10E-31 / 2) * (3.7^2 - .14^2) * 10E12 / 1.6 * 10E-19
V = -38.9 Volts
Since the proton is moving to lower potential energy it will gain velocity.
(The electric field is doing work on the proton - it takes work to move a positive
charge from infinity (V = 0) to a region with a positive potential)
Now just use the same equation to calculate the velocity of the proton
V = 1/2 m * (v1^2 - v2^2) / q
v2^2 = v1^2 - 2 * q * V / m   where V = -38.9 Volts
The origin is at the higher electric potential (for both electron and proton) but
the electron has a "lower" potential energy at the origin and loses KE while
the proton has a "higher" potential energy at the origin and gains KE.
The electric potential is in volts and does not depend on the sign of the charge.

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