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

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Homework Help Overview

The problem involves calculating the electric field produced by a uniformly charged nonconducting disk at a specific point along its axis. The charge distribution is given as -5.00 nC over a disk with a radius of 1.15 cm, and the point of interest is located 3.00 cm from the center of the disk.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the surface area element equation for the disk but encounters difficulties in calculating the electric field. Some participants question the applicability of the equation used, suggesting that it may only be valid for infinitely large plates. Others propose considering the electric field contribution from infinitesimally thin rings of charge and integrating over the disk.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the assumptions made regarding the equations. There is a suggestion to approach the problem by considering the disk as composed of concentric rings, which may lead to a more viable method for calculating the electric field.

Contextual Notes

Participants note that the original poster has uploaded an image of their attempt, indicating a collaborative effort to clarify the problem. There may be constraints related to the assumptions about the charge distribution and the geometry of the disk.

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.
 
... uploaded
 

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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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