How Does Charge Distribution Affect Electric Field Calculation on a Disk?

AI Thread Summary
To calculate the electric field produced by a uniformly charged nonconducting disk, one must consider the charge distribution and its effect on the field at a point along the axis. The charge of -5.00 nC is spread over a disk with a radius of 1.15 cm, and the electric field at a distance of 3.00 cm from the center needs to be determined. The surface area element for the disk is defined as sigma = Q / A, but this formula is only applicable for infinitely large plates. A more accurate approach involves using the formula E = kq/r^2 for an infinitely thin ring and integrating over the entire disk. This method accounts for the finite size of the disk and provides a correct calculation of the electric field.
mcaro
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Homework Statement



A charge of -5.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.15 cm. Find the magnitude of the electric field this disk produces at a point P on the axis of the disk a distance of 3.00 cm from its center.

Homework Equations



Surface area element for a disk [sigma]= Q / A

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The Attempt at a Solution



So I plugged the numbers in accordingly, had no luck.

I appreciate the help.
 

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I'm working on uploading a picture of my attempt.
 
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mcaro said:
Surface area element for a disk [sigma]= Q / A

That equation only applies for infinitely large plates. This disk is not infinitely large.

Try using E=kq/r^2 to get the electric field due to the charge on one infinitely thin ring surrounding the center of the disk. Then, integrate over the disk.
 
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