Electric field magnitude between two charged disks

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Homework Help Overview

The discussion revolves around calculating the electric field magnitude between two charged disks, specifically focusing on the scenario where one disk is positively charged and the other negatively charged. The problem involves understanding the electric field generated by finite-sized charged disks and how to apply the correct equations for this setup.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the appropriate equations for calculating the electric field from finite-sized disks, with one participant questioning the use of the equation for an infinitely charged plane. There is mention of needing to integrate to find the electric field for this configuration.

Discussion Status

The conversation indicates that participants are exploring the correct approach to the problem, with some guidance provided on the need for integration. There is acknowledgment of assumptions made regarding the parameters used in calculations, and one participant expresses a realization about a calculation error.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the resources they can use. There is an emphasis on understanding the underlying physics rather than simply obtaining the answer.

mateo710
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Homework Statement



Two 10-cm-diameter charged disks face each other, 17 cm apart. The left disk is charged to +40 nC and the right disk is charged to -40 nC.

a)What is the electric field, both magnitude and direction, at the midpoint between the two disks?
b) What is the force on a -1.0 nC charge placed at the midpoint?

So I have been trying this problem for hours now and am already maxed out on tries for a. I know how to do b but need the answer for a to get b.

Up until now I have been using E=(charge density of disk)/2(epsilon naught) however I am thinking now that this equation is only used for finding the electric field of an infinitely charged plane. So what equation am I supposed to use instead?

Actively idle,

Matt
 
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Hello, Matt.
mateo710 said:
Up until now I have been using E=(charge density of disk)/2(epsilon naught) however I am thinking now that this equation is only used for finding the electric field of an infinitely charged plane.
That's right. You'll need the equation for a finite-sized disk.
So what equation am I supposed to use instead?

This is a standard problem that requires integration. If you are using one of the standard textbooks, check to see if it's worked out there. Otherwise, we can guide you a bit in working it through.
 
TSny said:
Hello, Matt.

That's right. You'll need the equation for a finite-sized disk.


This is a standard problem that requires integration. If you are using one of the standard textbooks, check to see if it's worked out there. Otherwise, we can guide you a bit in working it through.

okay I did some reading and found E=η/2ε(1-(z/sqrt(z^2-R^2)))

now tell me if any of these assumptions are wrong
z=the distance of point charge from plate, so .085m
R=radius disk, so .05m
η=density charge, 5.092E-6 N/m^2

i solved using those values and still didnt come out right. do i need to multiply the E by 2 because of the two discs?
 
The total electric field is the sum of all the individuals. Since you have a parallel plate capacitor with a positive charge (outward field) on one plate and a negative charge (inward field) on the other plate what do you think the field will look like exactly in between the two?
 
Thanks I ended up getting it. Turns out I suck at working a calculator.
 

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