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.

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

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.

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) = 1.26 m/s

C. The electric field produced by a parallel plate capacitor is given by the formula

E = /0, so = E0 = q/A, so, finally

q = EA0 = (15 N/C)((0.058m)^2)(8.854*10^-12 F/m) = 4.47*10^-13 C

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