# Help - webassign due tonight - need some help

help -- webassign due tonight -- need some help

## Homework Statement

A parallel-plate capacitor has a plate area of 160 cm2 and a plate separation of 0.0400 mm.

A). Determine the capacitance.
which i got to be 3.54e-9 F

B). Determing the potential difference when the charge on the capacitor is 450.0 pC.
______V

C= ɛo A/d

## The Attempt at a Solution

i got part a using that equation, and i know that i should use it for part B, but i'm confused as to how to incorporate the 450pC....

Do you by chance go to NTHS?I have the exact same problem for my web assign except I have a 180 cm^2 and I need help getting that. For B I think you get the A answer and divide by the charge.

haha yea i do go to NTHS.

by the way what IS the charge??

Can you tell me how you got A.? for B you do C=Q/change of V, change of V you get by = -E(Change of d) because it's in a uniform electric field. The pC number, I think.
I was wrong on the previous post. it's asking for the potentical difference. I thought C, was the capacitors

Last edited:
for A you do ɛo A/d

ɛo is a constant, which is 8.85E-12
A is your centimeter number, but watch out cause it's squared so for me 160cm^2 = 160E-4

and d is your mm to meters, for me it's .04E-3

by the way i still cant get B

is the unit for potential difference in volts?

it's supposed to be in volts, but i honestly have no friggin idea what is C and Q and -E(change of d) ?

I got it! change of V = Q/C.
Q= the pC thing
C is the answer for a

Do you have
11. b/c
9.
or
3. a/b
by any chance?

Last edited:
it worked! awesome.

did you get 3b?? the angle thing?

I have no idea on how to do any of 3

thanks anyways