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Throughout space there is a uniform electric field in the -y direction of strength E = 450 N/C. There is no gravity. At t = 0, a particle with mass m = 5 g and charge q = -11 µC is at the origin moving with a velocity v0 = 35 m/s at an angle q = 25° above the x-axis.

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i found out that the magnitude of the force acting on the particle is 0.00495 N.

the question is this: at t = 5 s, what are the x- and y-coordinatesof the position of the particle?

x: 158.60 m, but i'm having trouble finding out what the y-coordinate would be, because of the electric field present in the -y direction.

i tried this:

the initial velocity in the y diction is 35 sin 25'.

then, v_y = 35sin25 - at, where a = F/m. i've already found F, and m is given. so, a = 0.99

then, v_y = 35sin25 - 0.99t. plug in t, i get 9.8416. but that's wrong.

what am i doing wrong?

thanks.

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# Homework Help: E-field of an ununiformly-charged rod

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