Parallel plate capacitor electric field problem

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Homework Help Overview

The problem involves a parallel-plate capacitor formed by two circular disks spaced 0.50 mm apart. The transfer of 1.8*10^9 electrons results in an electric field strength of 1.3*10^5 N/C, and the goal is to determine the diameters of the disks.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between charge, electric field, and area, with some attempting to manipulate equations to isolate variables. Questions are raised about how to derive the radius from the given equations and the implications of precision in online assignments.

Discussion Status

Some participants are exploring different equations and attempting substitutions to find relationships between variables. There is acknowledgment of confusion regarding the calculations and the definitions of terms, with no clear consensus reached on the approach to take.

Contextual Notes

Participants express frustration with the grading system of online assignments, which may affect their motivation and approach to solving the problem. There is also mention of revisiting foundational concepts in electricity to clarify understanding.

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Homework Statement



Two circular disks spaced 0.50 mm apart form a parallel-plate capacitor. Transferring 1.8*10^9 electrons from one disk to the other causes the electric field strength to be 1.3* 10^5 N/C. What are the diameters of the disks?

Homework Equations



F=k(q1)(q2)/r^2
E_capacitor= Q/(epsilon_0)*A


The Attempt at a Solution

 
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Look at your second equation. What can you find from it?
 
I'm doing the same problem.

All I can find from the second equation is that Q = Epsilon * A * E

Which doesn't really help because even if I break down A into pi * r^2 I still don't know r.. because if I knew r I would know d, which is what we're looking for.

I don't get how that helps.
 
Well I have another equation I was solving for earlier.

r = sqrt(Q/pi*eta)

I suppose I could substitute that into the equation set equal to Q and solve for Q.



Then maybe I could plug that Q from Q=Epsilon*A*E into the r = sqrt(Q/pi*eta) equation to find r.


Hmm. I can't stand this because of the precision required for these stupid online assignments, I get zero credit or full credit, none in between. Rewarded none for the effort I've put forth the same as someone who did nothing.

Oh well, computers give grades.
 
And now I end up with



Q=(Epsilon*E*Q) / eta



that makes no sense because if I go any further the Q's will cancel out *sigh*.. so I know I did something wrong.


Back to the drawing board.
 
Look at your second equation.

You know Q, you know ε0 and you know E. Can you find A? Once you know A can you find the diameter?
 
Last edited:
What is Q?

N electrons * Charge of electron?

I'm going back and re-reading the entire chapter leading up to this one in cased I missed something.

I'm completely lost when it comes to electricity and that's not like me.
 
thanks for the help. now if only webassign would stop lagging all the time and I could see if my answer is correct or not...
 
myxomatosii said:
What is Q?

N electrons * Charge of electron?

I'm going back and re-reading the entire chapter leading up to this one in cased I missed something.

I'm completely lost when it comes to electricity and that's not like me.

Yes, Q is the total charge of the N electrons.
 

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