Finding the electric field magnitude in a parallel plate compacitor.

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Homework Help Overview

The discussion revolves around calculating the electric field magnitude in a parallel plate capacitor, with a specific focus on the motion of an electron within the capacitor. The original poster describes their approach to solving the problem, including calculations related to the electron's speed, displacement, and forces acting on it. Additionally, there is a secondary question regarding the net torque on a rod with point charges in an external electric field.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the electric field by analyzing the motion of an electron and applying kinematic equations. Some participants suggest re-evaluating specific arithmetic steps and emphasize the importance of not rounding intermediate values. There are also inquiries about the method to find forces on a rod with fixed charges in an external electric field, with discussions on relevant equations.

Discussion Status

The discussion is active, with participants providing guidance on arithmetic corrections and clarifying the appropriate equations to use for the secondary question. There is an ongoing exploration of the concepts involved, particularly regarding the forces and torques related to the rod in the electric field.

Contextual Notes

Participants are navigating through the complexities of electric fields and forces, with some confusion about the relevance of certain equations. The original poster expresses uncertainty about their calculations and seeks clarification on the setup of the second problem involving the rod and external electric field.

nemisisnik
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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=volt + .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!
 

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nemisisnik said:
then:
x=volt + .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.
 
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?)
 
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 ?
 
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.)
 
ahh ok, i got it right, thank you so much for your help :)
 

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