Electric Fields of horizontal sheets

AI Thread Summary
The discussion focuses on calculating the electric field between two large horizontal sheets with equal but opposite surface charge densities. The user correctly identifies that the gravitational force on an oil droplet must be balanced by the electric force for it to remain suspended. They calculate the net charge of the droplet and attempt to find the electric field using the formula E = σ/(2ε0), which is applicable for a single plate. However, the user is informed that this expression does not account for the combined effect of both sheets, prompting a need to reconsider the electric field in the region between the two plates. The key takeaway is the importance of recognizing the contributions of both charged sheets to the overall electric field.
sdoug041
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Homework Statement



Two very large horizontal sheets are 4.25 cm apart and carry equal but opposite uniform surface charge densities of magnitude \sigma. You want to use these sheets to hold stationary in the region between them an oil droplet of mass 364 \mug that carries an excess of five electrons. Assume that the drop is in vacuum.

Homework Equations



E = FE / q

Fg = mg

E = \sigma / 2\epsilon0

\epsilon0 = 8.854 * 10 -12 C2/N*m2

The Attempt at a Solution



I calculated the Fg because I know the FE must be equal and opposite of gravity in order to be suspended.

I also calculated the net charge of the oil drop which was 5xe- (the charge of one electron) = 8.01 * 10-19 C.

I divided the FE by the net charge of the drop to find the Electric Field in N/C. I then multiplied E (electric field) by 2\epsilon0 to solve for \sigma but this is not the correct answer apparently.

Where have I went wrong?
 
Physics news on Phys.org
The expression E = σ/(2ε0) is the electric field due to a single plate. What is the electric field in the region between the two plates?
 

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