Finding the electric field magnitude in a parallel plate compacitor.

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SUMMARY

The discussion focuses on calculating the electric field magnitude in a parallel plate capacitor, where an electron with an initial speed of 9.42 x 10^6 m/s travels through a 2.00 cm long capacitor with a plate separation of 0.150 cm. The user initially calculated the acceleration as 8.87 x 10^15 m/s² and the force as 8.08 x 10^-15 N, leading to an electric field of 50524.3 N/C. However, the user questioned the accuracy of their calculations and sought clarification on the correct approach to determine the electric field and the torque on a rod with fixed charges in an external electric field.

PREREQUISITES
  • Understanding of electric fields and forces (E = F/q)
  • Familiarity with kinematic equations (x = Vo + 0.5at²)
  • Knowledge of torque and its calculation in electric fields
  • Basic principles of electrostatics, including point charges and electric field equations
NEXT STEPS
  • Review the calculation of electric fields in parallel plate capacitors using E = V/d
  • Study the principles of torque in electric fields, focusing on the formula τ = r × F
  • Learn about the effects of external electric fields on dipoles and charged rods
  • Explore the relationship between force, electric field, and charge in various configurations
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Students and professionals in physics, particularly those studying electromagnetism, electrical engineering, and anyone involved in solving problems related to electric fields and forces in capacitors and charged objects.

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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