Velocity of a particle in a parallel plate, electric field problem

1. Oct 30, 2008

chem1

1. The problem statement, all variables and given/known data

A particle with a mass of 2.0 x10^-5 kg and a charge of +2.0 microColoumbs is released in a (parallel plate) uniform horizontal electric field of 12 N/C

a) HOw far horizontally does the particle travel in 5.0 s?

b) What is the horizontal component of its velocity at that point?

c) If the plates are 5.0 cm on each side, how much charge is on each?

2. Relevant equations

I think I can use

a) F= qE = q(-Ey)
F= -qE= may
-qEy/m= ay but for x

b) x= xo + V0t + 1/2 at

V^2 = Vo^2 + 2a (x-xo)

c) E= 4(pi)kQ/ A => Q= EA/ 4(pi)k

3. The attempt at a solution

a) I assumed the above equation would be the same if applicable to x-axis, assuming the direction of the velocity is positive on the horizontal, I got

ax= qE/ m => a= (2.0 x10^-6) (12 ) / (2.0X10^-5) = 1.2 m/s^2

from that acceleration, I assumed, since the behaviour of the particle would be like a projectile motion since the electronegative field would attract the particle downward, the x-component of the velocity does not change, so v = vo then, I used the regular equation of kinematics to find x

x= xo + Vot + 1/2 at, since Vo and xo are assumed to be 0, I got

x= 0 + 0 + 1/2 (1.2 m/s^2) (0.5s)= 0.3 m

Am I doing and assuming the right thing??

b) Then I can get from the velocity equation Vx

V^2= Vo^2 + 2a (x-xo)
sqrt ((2 (1.2 m/s^2) (0.3 m)

=0.8485 m/s

c) According to the general formula, to get the charge on each plate would be

Q= EA/ 4(pi) k

Q= (12 N/C) ( 0.5 m)^2 / (4(pi) (8.99 x10^9)

= 2.6556x 10^ -13

2. Oct 31, 2008

Hootenanny

Staff Emeritus
Note that the question states that the electric field is horizontal. Therefore, the electric force will be acting horizontally and the weight of the charge will be acting vertically. Since the charge experiences a non-zero force in both the x and y directions, neither the x nor y component of the velocity will be constant.

Does that make sense?