Asked by Kid
An electron moving in the + x-direction has an initial speed of 3.7*10^6 m/s at the origin. Its speed is reduced to 1.4*10^5 m/s at x = 2cm. Calculate the potential difference between the origin and that point. Which point is at the higher potential?
My work and calculations:
-ΔU = ΔKE
-eΔV = (1/2)(m)(vf)^2 - (1/2)(m)(vi)^2
-(-1.602*10^-19)(ΔV) = (1/2)(9.11*10^-31kg)((1.4*10^5m/s)^2 - (3.7*10^6m/s)^2)
ΔV = -38.8693V = -39V
There is a higher potential at x = 0.
- Is all of this correct?
My work and calculations:
-ΔU = ΔKE
-eΔV = (1/2)(m)(vf)^2 - (1/2)(m)(vi)^2
-(-1.602*10^-19)(ΔV) = (1/2)(9.11*10^-31kg)((1.4*10^5m/s)^2 - (3.7*10^6m/s)^2)
ΔV = -38.8693V = -39V
There is a higher potential at x = 0.
- Is all of this correct?
Answers
Answered by
RZ-Digital-tech
Good.Because very easily simplified
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