I do not have time to run your numbers but your approach is correct.
constant Vx = 2.4*10^6 cos 38
so t = .015 /Vx
Vyi = 2.4*10^6 sin 38
y = Vyi t -(1/2)(qE/m) t^2 = 0 at x axis
A particle leaves the origin with a speed of 2.4 times 10^6 m/s at 38 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x = 1.5 cm if the particle is an electron.
I don't know what I am doing wrong on this problem. So I used the equation y(t)= y0 + v0t + ((1/2)qEt^2)/m
I rearranged it to solve for E in terms of t and got E= 2.733e-5/t then I plugged in the t I got from finding x=vxt=1.5cm
My answer came out to be 3.44e3 N/C
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