A particle with a mass of 2.0 × 10^-5 kg and a charge of +2.8 uC is released in

Question

A particle with a mass of 2.0 × 10^-5 kg and a charge of +2.8 uC is released in a (parallel plate) uniform horizontal electric field of 15 N/C.

Part A.
How far horizontally does the particle travel in 0.60 s ? Express your answer using two significant figures.

answer is= 0.38 m

Part B.
What is the horizontal component of its velocity at that point? Express your answer using two significant figures.

answer is= 1.3 m/s

Part C.
If the plates are 5.8 cm on each side, how much charge is on each? Express your answer using two significant figures.

I can't figure out the answer for this one. They want the answer in picocoloumbs though.

Explanation / Answer

A. The force that the charge experiences is F = qE = (2.8*10^-6 C)(15 N/C) = 4.2*10^-5 N So, the acceleration it undergoes is a = F/m = (4.2*10^-5 N)/(2*10^-5 kg) = 2.1 m/s^2 So, if it is released from rest, it's position at any time t is x(t) = (1/2)at^2 --> x(0.6s) = (1/2)(2.1 m/s^2)(0.6 s)^2 = 0.38 m B. Starting at rest, it's velocity at any time t is v(t) = at --> v(0.6 s) = (2.1 m/s)(0.6 s) = 1.3 m/s C. The electric field produced by a parallel plate capacitor is given by the formula E = s/e0, so s = Ee0 = q/A, so, finally q = EAe0 = (15 N/C)((0.058m)^2)(8.854*10^-12 F/m) = 4.47*10^-13 C = 0.447*10^-12 C = 0.447 pC

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