Find the maximum potential difference that can be applied to a capacitor

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 15K views
staticd
Messages
60
Reaction score
0

Homework Statement


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.

Homework 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)

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...
 
Physics news on Phys.org
Don't you need to know the dielectric strength of Teflon?
 
Donaldos said:
Don't you need to know the dielectric strength of Teflon?

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.
 
Donaldos said:
Don't you need to know the dielectric strength of Teflon?

staticd said:
That would be "k"-->k=2.1
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.