How Do I Calculate the Potential Difference in a Capacitor with a Given Charge?

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Homework Help Overview

The discussion revolves around calculating the potential difference in a parallel-plate capacitor given its charge and dimensions. The original poster has provided specific values for the capacitor's area and separation, as well as the charge, and is seeking assistance with the calculations involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula for capacitance and its application to find the potential difference. There are attempts to clarify the relationship between charge, capacitance, and potential difference. Some participants express confusion about the units and the variables involved.

Discussion Status

There is a mix of attempts to solve the problem, with some participants sharing their calculations and others asking for clarification on specific terms and concepts. While some guidance has been offered regarding the formulas, there is no explicit consensus on the approach to part B of the problem.

Contextual Notes

Participants are working under a time constraint due to an impending homework deadline. There are also variations in the problem setup among participants, which may affect their understanding and approach.

lettertwelve
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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...
 
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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 can't 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?
 
  • #10
I have no idea on how to do any of 3
 
  • #11
thanks anyways
 

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