What is the magnitude of charge inside the box

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To determine the magnitude of charge inside a box measuring 4.15 cm on each side, Gauss's law is applied using the measured electric field of 18.3 N/C. The calculation involves setting up the equation 18.3 = q/(2ε₀) to solve for the charge q. The total electric flux through the box's surfaces is calculated to find the total charge. Clarification is sought on whether the electric field is uniform on all sides or just two. The discussion concludes with acknowledgment of the guidance provided.
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?
 
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Got it

Got it Thanks Doc.
 

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