Finding the electric field magnitude in a parallel plate compacitor.

In summary, we discussed how to find the electric field magnitude for an electron entering a parallel plate capacitor. We also looked at a problem involving a long rod with point charges at each end and a constant external electric field, and determined how to find the net torque applied to the rod.
  • #1
nemisisnik
3
0
Problem:
The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is 9.42 x 10^6 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

What i did:

x=.5(Vo + V)t
.02m=.5(9.42 x 10^6 m/s + 9.42 x 10^6 m/s)t
solve for t=2 x 10^-9
then:
x=Vot + .5at^2
.0015m=0 + .5(a)(2 x 10^-9)^2
solve for a=8.87 x 10^15
then:
F=ma
F=(9.11 x 10^-31)(8.87 x 10^15)
F=8.08 x 10^-15
then:
E=F/q
E=8.08 x 10^-15/(1.6 x 10^-19)
E=50524.3 N/C

But that is the wrong answer, i was just wondering what i did wrong and how to correct myself.


In Addition i was wondering how to do this one?:
A long, thin rod (length = 3.0 m) lies along the x axis, with its midpoint at the origin. In a vacuum, a +7.0 C point charge is fixed to one end of the rod, and a -7.0 C point charge is fixed to the other end. Everywhere in the x, y plane there is a constant external electric field (magnitude 2.0 x 103 N/C) that is perpendicular to the rod. With respect to the z axis, find the magnitude of the net torque applied to the rod.

Thank you in advanced!
 

Attachments

  • Untitled.jpg
    Untitled.jpg
    4.4 KB · Views: 461
Physics news on Phys.org
  • #2
nemisisnik said:
then:
x=Vot + .5at^2
.0015m=0 + .5(a)(2 x 10^-9)^2
solve for a=8.87 x 10^15
Redo this arithmetic.

Also: Don't round off intermediate values (such as your value for t); wait until the last step.
 
  • #3
nemisisnik said:
In Addition i was wondering how to do this one?:
A long, thin rod (length = 3.0 m) lies along the x axis, with its midpoint at the origin. In a vacuum, a +7.0 C point charge is fixed to one end of the rod, and a -7.0 C point charge is fixed to the other end. Everywhere in the x, y plane there is a constant external electric field (magnitude 2.0 x 103 N/C) that is perpendicular to the rod. With respect to the z axis, find the magnitude of the net torque applied to the rod.
Start by finding the force on each end of the rod. Once you have the forces, you should be able to compute the torque about the z-axis. (What direction does the field point?)
 
  • #4
Doc Al said:
Start by finding the force on each end of the rod. Once you have the forces, you should be able to compute the torque about the z-axis. (What direction does the field point?)

How do you find the force on each end of the rod?
Should i just us the equation: E=kq/r^2 and then use E=F/q ?
 
  • #5
nemisisnik said:
Should i just us the equation: E=kq/r^2
No, that formula gives you the field from a point charge. Not relevant here, since you are given the field.
and then use E=F/q ?
That's the one you want. (F = Eq.)
 
  • #6
ahh ok, i got it right, thank you so much for your help :)
 

1. How do I calculate the electric field magnitude in a parallel plate capacitor?

To calculate the electric field magnitude in a parallel plate capacitor, you can use the formula E = V/d, where E is the electric field magnitude, V is the potential difference between the plates, and d is the distance between the plates. This formula assumes that the electric field is uniform between the plates.

2. What is the unit of measurement for electric field magnitude?

The unit of measurement for electric field magnitude is volts per meter (V/m) in the SI system. In other systems, it can also be measured in newtons per coulomb (N/C) or newtons per ampere (N/A).

3. How does the distance between the plates affect the electric field magnitude in a parallel plate capacitor?

The electric field magnitude is inversely proportional to the distance between the plates. This means that as the distance between the plates increases, the electric field magnitude decreases and vice versa. This relationship is expressed in the formula E = V/d, where d is the distance between the plates.

4. Can the electric field magnitude be negative in a parallel plate capacitor?

Yes, the electric field magnitude can be negative in a parallel plate capacitor. This occurs when the potential difference between the plates is negative. The direction of the electric field is always from the positive plate to the negative plate, regardless of the sign of the potential difference.

5. How does the potential difference between the plates affect the electric field magnitude in a parallel plate capacitor?

The electric field magnitude is directly proportional to the potential difference between the plates. This means that as the potential difference increases, the electric field magnitude also increases and vice versa. This relationship is expressed in the formula E = V/d, where V is the potential difference between the plates.

Similar threads

  • Introductory Physics Homework Help
2
Replies
58
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
299
  • Introductory Physics Homework Help
Replies
2
Views
855
  • Introductory Physics Homework Help
Replies
5
Views
789
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
228
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
625
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
948
Back
Top