Finding Dimensions of plates in Electric Field

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Homework Statement


The electric field between two circular plates of a capicator is changing at a rate of 1.5 x 10^6 V/m per Second. If the displacement current this instant is ID = 0.80 x 10^-8 A find the dimensions of the plates

Homework Equations




The Attempt at a Solution


The capacitance of a parallel plate capacitor is C = A*er*eo/t where A is the area of the plates, e is the relative permittivity of the dielectric, eo is the permeability of free space and t is the separation of the plates.

Q= V*C, where Q is the charge of the capacitor, V the voltage,
so we can write Q = A*E*er*eo, or Q/E = A*er*eo which implies (dQ/dt)/(dE/dt) = A*er*eo

I = dQ/dt, leading to I/(dE/dt) = A*er*eo = 0.80*10^-8/(1.5*10^6) = 5.33*10^-15

A = 5.33*10^-15/(er*eo)

If A is known the diameters of the circular plates could be determined, but A depends on er, and no information is given concerning this....I dont think I did this right.
 

Answers and Replies

  • #2
gneill
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I think you make the assumption that the relative permittivity is 1, that is, there is no dielectric in place since none was mentioned.

Otherwise you seem to be doing fine.
 
  • #3
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I think you make the assumption that the relative permittivity is 1, that is, there is no dielectric in place since none was mentioned.

Otherwise you seem to be doing fine.

So there are no dimensions ??
 
  • #4
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So there are no dimensions to the plate? That doesn't seem correct.
 
  • #5
gneill
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The plates will have dimensions of course. Square length units.

Note that ##ε_o## has units, as do the other quantities in your expression.
 
  • #6
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Yes so eo is 1.26 x10^ -6 m kg s ^-2 A^ -2.......than what is er? 1? What would the units be for that? And the units for 5.33x10^-15?
 
  • #7
gneill
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Yes so eo is 1.26 x10^ -6 m kg s ^-2 A^ -2.......than what is er? 1? What would the units be for that? And the units for 5.33x10^-15?
Relative permittivity is unitless, just a number.

You'll have to work out the units for your 5.33x10^-15 from the units of the values used to compute it, like dE/dt having units of V/m/s,
and Volts having equivalent units of J/C or ##kg~m^2 s^{-3} A^{-1}##, and so on.
 
  • #8
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If er is then 1....the solution comes out to 4.23x10^-9?
 
  • #9
gneill
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If er is then 1....the solution comes out to 4.23x10^-9?
What are the units? (seems like a pretty small area to me). Show your calculations.
 
  • #10
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A= 5.33x10^-15 / ( 1 x 1.26x10^-6 mkg s -2 A -2)
 
  • #11
gneill
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I'm not seeing units associated with your 5.33x10^-15 number, and I don't recognize your value of 1.26x10^-6 mkg s -2 A -2. You need to show your work and the values of any variables and constants.

I can see that you've divided the displacement current (C/s) by the rate of change of the electric field (V/m/s) to arrive at the 5.33x10^-15 number. So what are its units?

Would it help to know that equivalent units for ##\epsilon_o## are C/V/m ?
 
  • #12
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Units for the 5.33x10^-15 would be kg m^2 s^-3 A ^-2 m/s ??
 
  • #13
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I'm not seeing units associated with your 5.33x10^-15 number, and I don't recognize your value of 1.26x10^-6 mkg s -2 A -2. You need to show your work and the values of any variables and constants.

I can see that you've divided the displacement current (C/s) by the rate of change of the electric field (V/m/s) to arrive at the 5.33x10^-15 number. So what are its units?

Would it help to know that equivalent units for ##\epsilon_o## are C/V/m ?

I got 1.26x10^-6 m kg s^-2 A^-2 because thats the value for M naught?
 
  • #14
gneill
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I got 1.26x10^-6 m kg s^-2 A^-2 because thats the value for M naught?
You mean μo? That may be true, but there should not be a μo used here, there should be an εo which is a different constant.

μo crops up mostly when inductors and currents and magnetic fields are involved. εo comes into play when electric fields and capacitors are involved. It may be handy to remember that some alternate units for εo are F/m, and for μo are N/Amp2.
 
  • #15
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ahhh kk thank you, so eo would be 8.85 × 10-12 C2 N-1 m-2
 
  • #16
gneill
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ahhh kk thank you, so eo would be 8.85 × 10-12 C2 N-1 m-2
Sure.
 
  • #17
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I still dont understand where er is calculated though? As before you said it would just be a constant of 1?
 
  • #18
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No nevermind that cant be right, that gives me 4.72x10^-26...
 
  • #19
gneill
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I still dont understand where er is calculated though? As before you said it would just be a constant of 1?
Yes. There is no dielectric present so assume that the permittivity is just that of free space, ##\epsilon_o##.
 
  • #20
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So my final equation would be A= (5.33x10^-15kg m2 S-3 A-3 m/s)(1)(8.85x10^-12 C2N-1m-2) ?
 
  • #21
gneill
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You've missed a division (or multiplied where you should have divided). Go back to your first post and look at your last equation for A.
 
  • #22
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I get A = 6.02x10^-4 and I have no idea on the units now. I divided 5.33 x10^ -15 kg m2 s-3 A-2 m/s by 8.85x10^-12 C2 N-1 m-2
 
  • #23
gneill
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A big hint would be that it's an area, and you're working with fundamental units. So what are the units of area?
 
  • #25
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there are many different units for area, but this would break down into square meters?
 

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