Electron Force Between Parallel Plates

[ToffeeCake]

Homework Statement

[PLAIN]http://img214.imageshack.us/img214/6328/physics2.jpg
An electron enters the lower left side of a parallel plate capacitor and exists precisely at the upper right side (just clearing the the upper plate). The initial velocity of the electron is 7x10^6 m/s parallel to the plates (see diagram). The capacitor is 2 cm long and its plates are separated by 0.150 m. Assume the electric field is uniform at every point between the plates and find its magnitude.

The correct answer is supposed to be 2.09x10^5 N/C.

Homework Equations

Electric Force Equation

The Attempt at a Solution

Since E= F/q, I attempted to find F by stating F=ma. To find acceleration, I said v=d/t thus t=d/v. Then a=v/t. I plugged this "a" into F=ma and used the mass of an electron to calculate force and then divided by the electron's charge, but I didn't get the right answer. Any suggestions?

 

[collinsmark]

Hello ToffeeCake,

Welcome to Physics Forums!

Since E= F/q, I attempted to find F by stating F=ma. To find acceleration, I said v=d/t thus t=d/v. Then a=v/t. I plugged this "a" into F=ma and used the mass of an electron to calculate force and then divided by the electron's charge, but I didn't get the right answer. Any suggestions?

I just worked the problem and got 2.09 x 10⁵ N/C.

I find it very odd that the capacitors plates are 2 cm wide, yet separated by a distance of 15 cm. That makes the approximation of the "electric field is uniform at every point between the plates" rather unrealistic.

But whatever the case, working out the problem as it is stated (and applying the approximation as directed) produces the answer of 2.09 x 10⁵ N/C. So you'll have to show us your work and we can help you figure out what went wrong.

 

[ToffeeCake]

My work is as shown below:

Mass of Electron: 9.11 x 10^-31 kg
Charge of Electron: 1.6 x 10^-19 C

E= F/q
F= ma
v= d/t
t= d/v= 0.02/(7x10^6)= 2.86 x 10^-9 seconds
a= v/t= (7x10^6)/(2.86x10^-9)= 2.45 x 10^15 m/s/s
F= ma= (9.11x10^-31)*(2.45x10^15)= 2.23 x 10^-15 N
E= F/q = (2.23x10^-15)/(1.6x10^-19)= 13949.7 N/C

 

[cdotter]

I'm taking physics 2 right now so I'm not entirely sure I'm correct, but I think you're wrong here:

a= v/t= (7x10^6)/(2.86x10^-9)= 2.45 x 10^15 m/s/s.

The time you calculated to travel .02 m is correct. In this time the electron travels 0.15 m from the bottom to the top. You can use a kinematics equation to find acceleration. The rest your work will be correct when you substitute the correct acceleration in.

 

[collinsmark]

My work is as shown below:

Mass of Electron: 9.11 x 10^-31 kg
Charge of Electron: 1.6 x 10^-19 C

E= F/q
F= ma
v= d/t
t= d/v= 0.02/(7x10^6)= 2.86 x 10^-9 seconds

So far so good.

a= v/t= (7x10^6)/(2.86x10^-9)= 2.45 x 10^15 m/s/s

a ≠ v/t.

You can't just do that. If you happened to know the initial and final velocity components of a given direction (such as the y-direction) you could calculate the acceleration in that direction, assuming it is uniform, via
a_y = (v_yf - v_yi)/t
But the y-component of the final velocity is not given. So this method doesn't apply here.

But there is a method that does. You know that F = ma, and you know that E = F/q. Combine these equations into one.

You also have another kinematics equation for uniform acceleration that relates displacement, acceleration and time. Combine that one into your existing combination too. That's all you need.

 

[ToffeeCake]

You also have another kinematics equation for uniform acceleration that relates displacement, acceleration and time. Combine that one into your existing combination too. That's all you need.

So this is what I ended up doing:
d= v_it + ½at²
a= 2(d-v_it) / t²
= [2(0.02-(7x10⁶)(2.86x10^-9))] / ((2.86x10^-9)²)
= -4.89 x 10¹² m/s/s
^ My first question is whether this should be a negative acceleration.. which I don't quite understand

E= ma/q
E= [(9.11x10^-31)(-4.89x10¹²)]/(1.6x10^-19)= -27.84 N/C

 

[ToffeeCake]

Nevermind! I got it :) I was using horizontal displacement and velocities instead of vertical.
It slipped my mind that that vi_y= 0 m/s.

Anyways, thank you! I figured it out this time