Help - webassign due tonight - need some help

  • #1
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


Homework Equations



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....
 

Answers and Replies

  • #2
7
0
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.
 
  • #3
haha yea i do go to NTHS.

by the way what IS the charge??
 
  • #4
7
0
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:
  • #5
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
 
  • #6
7
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is the unit for potential difference in volts?
 
  • #7
it's supposed to be in volts, but i honestly have no friggin idea what is C and Q and -E(change of d) ?
 
  • #8
7
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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:
  • #9
it worked! awesome.

did you get 3b?? the angle thing?
 
  • #10
7
0
I have no idea on how to do any of 3
 
  • #11
thanks anyways
 

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