Electric Field: Why is This Statement Wrong?

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Homework Help Overview

The discussion revolves around the concept of electric fields and Gauss's law, specifically addressing the influence of external charges on the electric field at a Gaussian surface. Participants are exploring the relationship between enclosed charge and the electric field generated by charges outside the surface.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the assertion that the electric field on a Gaussian surface is unaffected by external charges, with some exploring the implications of Gauss's law and the conditions under which it applies. There is a focus on understanding how the electric field behaves in relation to both enclosed and external charges.

Discussion Status

The discussion is active, with participants sharing insights and clarifying concepts. Some have expressed frustration, indicating a need for deeper understanding. There is an acknowledgment of the changing nature of the electric field distribution on the Gaussian surface as external charges are considered.

Contextual Notes

Participants are grappling with the implications of Gauss's law and the conditions necessary for its application, particularly in scenarios involving electric dipoles and varying Gaussian surface sizes.

yti1211
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y is this statement wrong?

The electric field on a gaussian surface is generally not influenced by charge that is not enclosed by the surface.

ps: isn't the electric field proportional only to the ENCLOSED CHARGE, but not the outside charge?? :confused:

This is getting me so fRustrated
 
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Gauss law is a law of electric flux, not E-field. If the situation is symmetric enough, you may extract the value of the E-field from it. Consider a Gaussian surface enclosing an electric dipole. Are you able to determine the E-field at any point on the surface? Now imagine the distance between the positive and negative charges increasing along with the size of the Gaussian surface. What is the E-field at any point on the surface?
 
hmm, first, thanks Defender! I no that u need a constant E throughout the Gaussian surface in order to extract E from the integral. but how do you explain why the E- field on Gaussian surface is affected by a charge outside of the surface?? because the ENCLOSED Q is the key right? or do u mean the size of the Gaussian surface is changing when a charge is placed outside of it?

THXXX :)
 
ohh, I think I got it now! So the distribution of E field is changing on the Gaussian surface, but the total E dA is the same. yAy
 

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