Electric field magnitude between two charged disks

  • Thread starter mateo710
  • Start date
  • #1
mateo710
4
0

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
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
13,659
3,840
Hello, Matt.
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.
 
  • #3
mateo710
4
0
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?
 
  • #4
Bhumble
157
0
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?
 
  • #5
mateo710
4
0
Thanks I ended up getting it. Turns out I suck at working a calculator.
 

Suggested for: Electric field magnitude between two charged disks

Replies
1
Views
346
Replies
2
Views
303
Replies
58
Views
2K
Replies
9
Views
377
Replies
12
Views
497
  • Last Post
Replies
1
Views
261
Top