# Finding Dimensions of plates in Electric Field

1. Nov 25, 2015

### lloyd21

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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.

2. Nov 25, 2015

### Staff: Mentor

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. Nov 28, 2015

### lloyd21

So there are no dimensions ??

4. Nov 28, 2015

### lloyd21

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

5. Nov 28, 2015

### Staff: Mentor

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. Nov 28, 2015

### 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?

7. Nov 28, 2015

### Staff: Mentor

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. Nov 28, 2015

### lloyd21

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

9. Nov 28, 2015

### Staff: Mentor

What are the units? (seems like a pretty small area to me). Show your calculations.

10. Nov 28, 2015

### lloyd21

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

11. Nov 28, 2015

### Staff: Mentor

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. Dec 1, 2015

### lloyd21

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

13. Dec 1, 2015

### lloyd21

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

14. Dec 1, 2015

### Staff: Mentor

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. Dec 1, 2015

### lloyd21

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

16. Dec 1, 2015

### Staff: Mentor

Sure.

17. Dec 1, 2015

### lloyd21

I still dont understand where er is calculated though? As before you said it would just be a constant of 1?

18. Dec 1, 2015

### lloyd21

No nevermind that cant be right, that gives me 4.72x10^-26...

19. Dec 1, 2015

### Staff: Mentor

Yes. There is no dielectric present so assume that the permittivity is just that of free space, $\epsilon_o$.

20. Dec 1, 2015

### 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) ?