Find the electric field at a point away from the charged cylinder's axis

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Homework Help Overview

The discussion revolves around finding the electric field at a point outside a charged conducting cylinder. The problem involves understanding the behavior of electric fields in relation to conductors and the distribution of charge.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of having no charge inside the conducting cylinder and the resulting electric field. There are attempts to clarify the location of the point in relation to the conducting material and the charge distribution.

Discussion Status

Some participants have provided insights regarding the charge on the conducting shell and its implications for the electric field. There appears to be a productive exploration of the concepts involved, although no consensus has been reached on the final interpretation.

Contextual Notes

Participants note that the point of interest is located 10 cm from the axis of the cylinder, which is beyond the conducting shell. The charge on the conducting shell is specified as negative, which may influence the discussion further.

Fatima Hasan
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Homework Statement


screenshot_56.png


Homework Equations



##E=\frac{kQ}{r^2}##

3. The Attempt at a Solution

##Q_{enclosed} = 0 ## , because there is no charge inside the conducting cylinder.
E = 0
Can anyone check my answer please ?
 

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Fatima Hasan said:
because there is no charge inside the conducting cylinder.
The point in question is not inside the conducting material. It's at a point 10 cm from the axis, well beyond the conducting shell.
 
Doc Al said:
The point in question is not inside the conducting material. It's at a point 10 cm from the axis, well beyond the conducting shell.
rac%7B20*10%5E%7B-9%7D+12*10%5E%7B-9%7D%7D%7B2%5Cpi%20*8.85*10%5E%7B-12%7D*0.1%7D%20%3D%205.png
 

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  • rac%7B20*10%5E%7B-9%7D+12*10%5E%7B-9%7D%7D%7B2%5Cpi%20*8.85*10%5E%7B-12%7D*0.1%7D%20%3D%205.png
    rac%7B20*10%5E%7B-9%7D+12*10%5E%7B-9%7D%7D%7B2%5Cpi%20*8.85*10%5E%7B-12%7D*0.1%7D%20%3D%205.png
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Note that the charge on the conducting shell is negative.
 
Doc Al said:
Note that the charge on the conducting shell is negative.
7B20*10%5E%7B-9%7D-12*10%5E%7B-9%7D%7D%7B2%5Cpi%20%5Ccdot%208.85*10%5E%7B-12%7D*0.1%7D%20%3D%201.png
 

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  • 7B20*10%5E%7B-9%7D-12*10%5E%7B-9%7D%7D%7B2%5Cpi%20%5Ccdot%208.85*10%5E%7B-12%7D*0.1%7D%20%3D%201.png
    7B20*10%5E%7B-9%7D-12*10%5E%7B-9%7D%7D%7B2%5Cpi%20%5Ccdot%208.85*10%5E%7B-12%7D*0.1%7D%20%3D%201.png
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Now you've got it.
 
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