Finding Dimensions of plates in Electric Field

[lloyd21]

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 don't think I did this right.

 

[gneill]

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.

 

[lloyd21]

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 ??

 

[lloyd21]

So there are no dimensions to the plate? That doesn't seem correct.

 

[gneill]

The plates will have dimensions of course. Square length units.

Note that $ε_o$ has units, as do the other quantities in your expression.

 

[lloyd21]

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?

 

[gneill]

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.

 

[lloyd21]

If er is then 1...the solution comes out to 4.23x10^-9?

 

[gneill]

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.

 

[lloyd21]

A= 5.33x10^-15 / ( 1 x 1.26x10^-6 mkg s -2 A -2)

 

[gneill]

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 ?

 

[lloyd21]

Units for the 5.33x10^-15 would be kg m^2 s^-3 A ^-2 m/s ??

 

[lloyd21]

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 that's the value for M naught?

 

[gneill]

I got 1.26x10^-6 m kg s^-2 A^-2 because that's the value for M naught?

You mean μ_o? That may be true, but there should not be a μ_o used here, there should be an ε_o which is a different constant.

μ_o crops up mostly when inductors and currents and magnetic fields are involved. ε_o comes into play when electric fields and capacitors are involved. It may be handy to remember that some alternate units for ε_o are F/m, and for μ_o are N/Amp².

 

[lloyd21]

ahhh kk thank you, so eo would be 8.85 × 10-12 C2 N-1 m-2

 

[gneill]

ahhh kk thank you, so eo would be 8.85 × 10-12 C2 N-1 m-2

Sure.

 

[lloyd21]

I still don't understand where er is calculated though? As before you said it would just be a constant of 1?

 

[lloyd21]

No nevermind that can't be right, that gives me 4.72x10^-26...

 

[gneill]

I still don't 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, $\epsilon_o$.

 

[lloyd21]

So my final equation would be A= (5.33x10^-15kg m2 S-3 A-3 m/s)(1)(8.85x10^-12 C2N-1m-2) ?

 

[gneill]

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.

 

[lloyd21]

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

 

[gneill]

A big hint would be that it's an area, and you're working with fundamental units. So what are the units of area?

 

[lloyd21]

square meters

 

[lloyd21]

there are many different units for area, but this would break down into square meters?

 

[gneill]

there are many different units for area, but this would break down into square meters?

Yes. All the other units that you've been using in the problem were scaled to the base units of kg, m, second,... so a result that is an area will have the units of square meters.

 

[lloyd21]

isnt A = 6.02x10^-4 m^2 too small for dimensions of a plate? Also, when i got my 5.33 x 10^-15 kg m2s-3A-2 m/s by dividing 0.80x10^8A by 1.5x10^6 vms, was that correct or should I have multiplied them instead?

 

[gneill]

Dividing is correct. It boils down to :

$Area = \frac{ID}{\epsilon_o \frac{dE}{dt}}$

The area is not too small. How could you judge? Without knowing the plate separation you can't say much about the capacitance.
If it helps, that area is the same as about 600 mm².

 

[lloyd21]

Would my answer still be correct if I leave it as A = 6.02x10^-4 m^2?

 

[gneill]

Would my answer still be correct if I leave it as A = 6.02x10^-4 m^2?

Yes. I just gave the mm² conversion so you might gain a different perspective on what the value represents.