Charge to Mass Ratio Calculation for Object

[koolmerliner]

Homework Statement

A small object with mass m, charge q, and initial speed v0 = 6.00×10^3 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm(Figure 1) . The electric field between the plates is directed downward and has magnitude E = 600 N/C . Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance d = 1.35 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance.

Part A
Calculate the object's charge-to-mass ratio, q/m.

Homework Equations

E = F/q
E=kq/r^2
kinematics

The Attempt at a Solution

I solved for q by using E = kq/r^2[/B]
Er^2/k = q
(600 N/C)(.26 m)^2 / 9x10^9 Nm^2/C^2 = q
q = 4.51 * 10^-9 C

My plan for mass is
Eq = F = ma
Eq/a = m

Using Kinematics
Solving for time t
x - x0 =1/2(v0x + vx)t
2(x-x0)/(v0x + vx) = t
2(.56m)/(2 * 6 *10^3 m/s) = t
t = 9.33*10^-5 s

Solving for Acceleration y-axis ay
y = y0 + v0yt + (1/2)ayt^2
2(y - y0 - v0yt)/t^2 = ay
2(.56m)/t^2 = ay
ay = 3127478 m/s

Eq/a = m
m = 8.668 * 10^-13 kg

q/m = 5203 wrong answer

 

[haruspex]

2(.56m)/t^2 = ay

The acceleration only applies while between the plates. Break the problem into two parts.

 

[kuruman]

And while you do what haruspex suggested, consider that the E-field in which the particle travels is not kq/r^2, which is the field generated by a point charge, but a uniform field of 600 N/C generated by external charges on the plates.

 

[koolmerliner]

I tried solving for Vy(in plates) which is equal to V0y and Vy (outside the plates)
y-y0 = 1/2(2Vy)t
(-0.0135m)/t = Vy
t = 5 * 10^-5 s (outside the from 26 to 56) calculated wrong in post
Vy = (-0.0135m)/t = -270 m/s

Now solving for t (x=0 to x=.26m)
(x-x0) = (1/2)(v0x + v0x)t
(.26m)/(6.00 * 10^3 m/s) = t
t = 4.33*10^-5 s

ay = (-270m/s)/ t
ay = 6235565 m/s^2

 

[kuruman]

A "previous question" is often an unreliable source. Here you don't have a line of charge, just a moving charge in a uniform electric field. The electric field in the line of charge question does not have the same dependence on space coordinates (it looks different if you draw field lines in space) as a uniform electric field, which is what you have here. Therefore, it is incorrect to use the line-of-charge of expression where it does not apply and expect to make sense out of it.

I have not done the math yet to see if your number for ay is correct. Assuming that it is, can you find an expression for ay in terms of the charge q, the external electric field E and the mass m?

 

[koolmerliner]

A "previous question" is often an unreliable source. Here you don't have a line of charge, just a moving charge in a uniform electric field. The electric field in the line of charge question does not have the same dependence on space coordinates (it looks different if you draw field lines in space) as a uniform electric field, which is what you have here. Therefore, it is incorrect to use the line-of-charge of expression where it does not apply and expect to make sense out of it.

I have not done the math yet to see if your number for ay is correct. Assuming that it is, can you find an expression for ay in terms of the charge q, the external electric field E and the mass m?

ay = qE/m

 

[kuruman]

Right. Now look at the equation. What quantity are you trying to find and how can you write it in terms of known quantities?