What is the magnitude of charge inside the box

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Homework Help Overview

The discussion revolves around determining the magnitude of charge inside a box based on electric field measurements. The subject area is electrostatics, specifically applying Gauss's law to relate electric fields and charge distributions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Gauss's law, with one suggesting a specific setup for calculating charge based on the electric field measurement. Questions arise regarding the uniformity of the electric field across the box's sides.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem and the application of Gauss's law. Some guidance has been provided regarding calculating flux and charge, but no consensus has been reached on the specifics of the setup.

Contextual Notes

Participants are working within the constraints of the problem as posed, including the dimensions of the box and the electric field measurement. There is a hint provided, but no further details on assumptions or additional information have been mentioned.

eltel2910
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have a box that is 4.15 cm on each side. Using a probe I measure on two opposite sides of the box and find a nearly uniform, inward-oriented, perpendicular component of the field with the magnitude of 18.3 N/C. What is the magnitude of charge inside the box?
 
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Hint: Use Gauss's law.
 
like this?

so do I set it up like: 18.3 = q/(2Eo). Solving for q and then that would give me one unit and then I would multiply by the volume of the cube (.00414^3)?
 
To find the charge within the box using Gauss's law, first calculate the flux through each side of the box. The total flux will allow you to calculate the total charge. See this for more on http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html" .

Is the field uniformly inward on all sides of the box? Or just two?
 
Last edited by a moderator:
Got it

Got it Thanks Doc.
 

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