Parallel plate capacitor problem

[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 ?

 

[LowlyPion]

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 ?

How will the dielectric strength constrain the distance separation?

Then solve for A.

 

[Apprentice123]

How will the dielectric strength constrain the distance separation?

Then solve for A.

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

 

[LowlyPion]

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

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?

 

[Apprentice123]

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?

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

It is related to d and A again

 

[Apprentice123]

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

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

It is ?

 

[LowlyPion]

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

It is related to d and A again

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.

 

[Apprentice123]

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.

Thank you very much!