# Parallel plate capacitor problem

1. Jun 30, 2009

### Apprentice123

A parallel plate capacitor is formed using a material whose dielectric constant is 3,00 and whose dielectric strength is 2x10^{8} V/m. The capacitance is 0,25x10^{-6}F and the capacitor must withstand a potential difference maximum of 4000V. Find the minimum area of the plates of the capacitor

I have

C = (E. * A * k)/d

How to calculate d ?

2. Jun 30, 2009

### LowlyPion

Re: Capacitor

How will the dielectric strength constrain the distance separation?

Then solve for A.

3. Jun 30, 2009

### Apprentice123

Re: Capacitor

the dielectric strength (2x10^8 V/m) is the separation of the plates?

4. Jun 30, 2009

### LowlyPion

Re: Capacitor

No. That's a property of the material if it was 1 meter thick.

How thick does it need to be to yield a 4000 V rating?

5. Jun 30, 2009

### Apprentice123

Re: Capacitor

V = (q*d)/(E.*A*k)

It is related to d and A again

6. Jun 30, 2009

### Apprentice123

Re: Capacitor

d = 4000 / 2x10^8 = 2x10^{-5}

A = (C*d)/(E.*k) = 1,88x10^{-7}

It is ?

7. Jun 30, 2009

### LowlyPion

Re: Capacitor

Like I said, find d.

For this problem minimum d is the thickness required to provide a 4000 V separation between the plates. That means that with a dielectric strength of 2*108 you need a thickness of 4000/2*108 2*10-5 m.

Then apply that to Area calculation.

Edit: I see you have in your next post. So yes. That looks like the right way to do it.

8. Jun 30, 2009

### Apprentice123

Re: Capacitor

Thank you very much!