Electric Field and Charge of a Capacitor

  • Thread starter Squall
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  • #1
Squall
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Homework Statement


In a cathode-ray tube (CRT), an electron travels in a vacuum and enters a region between two "deflection" plates which have equal and opposite charges. The dimensions of each plate are L = 13 cm by d = 4 cm, and the gap between them is h = 2.5 mm.

During a 0.001 s interval while it is between the plates, the change of the momentum of the electron is < 0, 5.60e-17, 0 > kg m/s.

What is the electric field between the plates?

What is the charge (both magnitude and sign) of the upper plate?
E is pointing upwards towards the upper plate from the bottom one.

Homework Equations


Impulse = Force * Time
F= q*Enet
E= (Q/A)/(permittivity constant) "per"


The Attempt at a Solution


F = <0,5.6e-17,0>(kg*m/s) / .001(s)
Enet = F/Q
Enet = (Q/A)/"per"
F/Q=(Q/A)/"per"
Q=sqrt("per"*F*A)
Q=sqrt(8.854e-12(C^2/Nm^2) * <0,5.6e-14,0>(N) * .0052(m^2)
Q=5.077e-14C

Well this is my attempt to find the charge and it seems I am doing something wrong, can you please point me in the right direction.

Thank You
 

Answers and Replies

  • #2
gneill
Mentor
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2,886
Calculate the field strength.
With the field strength and the geometry given, calculate the potential difference (voltage) across the plates.
Calculate the capacitance of the plate arrangement.
With the capacitance and the voltage across it, find the charge.
 
  • #3
Squall
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I am sorry but can you tell me how I would go about calculating the strength of the field. I assumed you would have to know the charge first to figure out the strength.
 
  • #4
Squall
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Tell me if I'm headed in the right direction since we are dealing with an electron I know the charge therefore should be able to figure out the electric field using the F=qE formula.
|F|= 5.6e-14N
q= -1.6e-19C
E=350,000 N/C
 
  • #5
Squall
53
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Thank You, I figured it out
 

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