# Homework Help: Find the maximum potential difference that can be applied to a capacitor

1. Oct 8, 2009

### staticd

1. The problem statement, all variables and given/known data
Determine a) the capacitance and b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.75 cm^2 and plate separation of .04 mm.
2. Relevant equations

A=.0175
d=.04e-3
k=2.1
C=k(epsilon-not)(A/d)
Vmax=E*d
Qmax=C*(delta)Vmax=C(Emax*d)

3. The attempt at a solution

C=2.1*8.85e-12*(.0175/.04e-3)=8.13e-9 F

If Vmax=E*d, how do I solve for E to find Vmax?

Not really sure where to go with this...
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 8, 2009

### Donaldos

Don't you need to know the dielectric strength of Teflon?

3. Oct 9, 2009

### staticd

That would be "k"-->k=2.1

I guess I need to know how to determine what the electric field is in between the plates, as a function of the dialectric.

4. Oct 9, 2009

### Redbelly98

Staff Emeritus
k is the dielectric constant. Dielectric strength is the maximum electric field that a material can withstand.

Chances are you were provided with dielectric strength values -- probably in your textbook. Check the relevant section, or try looking in the book's index.