Electric Field Between Two Disks

[k8thegr8]

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

E_disk = (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!

 

[vanhees71]

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

 

[vela]

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

E_disk = (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 E_disk?

 

[k8thegr8]

Are you sure about your formula for E_disk?

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 E_disk, 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.

 

[HalfLostAndSearching]

I can confirm that the answer of 3.9 x 104 N/C is correct. The Electric Field of a Disk formula you used is the correct formula to use in this scenario. The reason for doubling the answer is because the electric field is a vector quantity, meaning it has both magnitude and direction. In this case, the electric fields from the two disks are in opposite directions, so they will cancel each other out to some extent at the midpoint between the disks. Doubling the answer takes into account the vector nature of the electric field and gives the net result. I would suggest practicing more problems like this to become more comfortable with using the formula and understanding the concept of vector addition in electric fields. Good luck on your exam!