Electric Field Between Two Disks

AI Thread Summary
The discussion centers on calculating the electric field at the midpoint between two charged disks, one negatively charged at -50 nC and the other positively charged at +50 nC, separated by 29 cm. The correct electric field magnitude is stated as 3.9 x 10^4 N/C. Participants express confusion over the formula for the electric field of a disk, with one user attempting to apply the formula but unsure about the correct approach, especially regarding whether to double the result for two disks. There is frustration over discrepancies between online resources and the professor's notes. Clarification on the formula and its application is sought for exam preparation.
k8thegr8
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Homework Statement



Two 10-cm-diameter charged disks face each other, 29cm apart. The left disk is charged to -50 nC and the right disk is charged to +50 nC. What is the magnitude of the electric field E⃗ at the midpoint between the two disks?

The answer is 3.9 x 104 N/C


Homework Equations



Edisk = (KQ/R2)[1 - z/√(z2 + R2)]
Where
Q = Charge of the disk in Coulombs = 5 x 10-8 C
R = Radius of the disk in meters = 0.05m
Z = Distance from the disk in meters = 0.145m


The Attempt at a Solution



This seems like a simple plug-and-chug to me. I've tried using the Electric Field of a Disk formula above and I've tried doubling my answer to accommodate for the two plates. Can anyone else hit that magic number, and if so, how did you do it? I'd like to know for an exam coming up.

Thanks!
 
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Don't forget that you have two disks that are separated (by 29 cm)!
 
k8thegr8 said:

Homework Statement



Two 10-cm-diameter charged disks face each other, 29cm apart. The left disk is charged to -50 nC and the right disk is charged to +50 nC. What is the magnitude of the electric field E⃗ at the midpoint between the two disks?

The answer is 3.9 x 104 N/C


Homework Equations



Edisk = (KQ/R2)[1 - z/√(z2 + R2)]
Where
Q = Charge of the disk in Coulombs = 5 x 10-8 C
R = Radius of the disk in meters = 0.05m
Z = Distance from the disk in meters = 0.145m


The Attempt at a Solution



This seems like a simple plug-and-chug to me. I've tried using the Electric Field of a Disk formula above and I've tried doubling my answer to accommodate for the two plates. Can anyone else hit that magic number, and if so, how did you do it? I'd like to know for an exam coming up.

Thanks!
Are you sure about your formula for Edisk?
 
vela said:
Are you sure about your formula for Edisk?

You know, I'm beginning to think that my problem lies there. As I look around the internet, it seems like everyone has their own personal formula for Edisk, such as this website which essentially multiplies my formula by 2.

Now that's not to say I'm not peeved that my professor picked a homework website that didn't jive with his in-class notes, but that's a separate issue.
 
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