Electric Fields of horizontal sheets

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SUMMARY

The discussion centers on calculating the electric field between two large horizontal sheets with equal but opposite surface charge densities, specifically for an oil droplet with a mass of 364 µg and a net charge of 8.01 x 10-19 C. The user correctly identifies that the electric force (FE) must balance the gravitational force (Fg) for the droplet to remain suspended. However, the user incorrectly applies the formula E = σ/(2ε0) for a single plate instead of considering the combined effect of both plates, which results in an electric field of E = σ/ε0 in the region between the sheets.

PREREQUISITES
  • Understanding of electric fields and forces
  • Familiarity with Coulomb's law and charge interactions
  • Knowledge of the concept of surface charge density
  • Basic principles of electrostatics and gravitational forces
NEXT STEPS
  • Review the concept of electric fields produced by multiple charged plates
  • Study the derivation of the electric field between parallel plates
  • Learn about the relationship between charge, mass, and force in electrostatics
  • Explore applications of electric fields in real-world scenarios, such as capacitors
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding electrostatics, particularly in the context of charged objects and electric fields between parallel plates.

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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