desibabu90
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[SOLVED] Electric Field Problem
At some instant the velocity components of an electron moving between two charged parallel plates are Vx = 1.5E5 m/s and vy = 1.8E3 m/s. Suppose that the electric field between the plates is given by E = (140 N/C) j.
(a) What is the electron's acceleration in the field?
(b) What is the electron's velocity when its x coordinate has changed by 3.6 cm?
I got the first answer doing ---- a = (e/m) E.
so the answer came out to be -2.45E13.
For b. you have to find the x portion for the velocity and the y portion of the veloctity
i found the x, it just stays the same and for the y:
Kinematics:
D = v0^2 - v^2 / (2a)
2aD = V0^2 - V^2
V = sqrt(v0^2 + 2aD)
V = sqrt(v0^2 + 2aD)
Please help
Homework Statement
At some instant the velocity components of an electron moving between two charged parallel plates are Vx = 1.5E5 m/s and vy = 1.8E3 m/s. Suppose that the electric field between the plates is given by E = (140 N/C) j.
(a) What is the electron's acceleration in the field?
(b) What is the electron's velocity when its x coordinate has changed by 3.6 cm?
Homework Equations
I got the first answer doing ---- a = (e/m) E.
so the answer came out to be -2.45E13.
For b. you have to find the x portion for the velocity and the y portion of the veloctity
i found the x, it just stays the same and for the y:
Kinematics:
D = v0^2 - v^2 / (2a)
2aD = V0^2 - V^2
V = sqrt(v0^2 + 2aD)
The Attempt at a Solution
V = sqrt(v0^2 + 2aD)
Please help