Electric Flux Density: 4.5Xe + .50MV/m

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Discussion Overview

The discussion revolves around calculating the electric displacement flux density given a specific electric field and relative permittivity. The focus is on the application of relevant equations and the interpretation of units in the context of a homework problem.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant presents the formula for displacement flux density, D=ε0εrE, and attempts to calculate it using given values for Xe and E.
  • Another participant clarifies that the answers are provided in nC/m², prompting a reevaluation of the units used.
  • A subsequent post suggests that the correct units for the answer should actually be μC/m².
  • There is a proposal that the calculated value should be 24.338 μC/m², though this is not confirmed as correct.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct units for the answer, with differing views on whether it should be in nC/m² or μC/m².

Contextual Notes

There are unresolved issues regarding the conversion of units and the accuracy of the calculated displacement flux density. The discussion does not clarify the assumptions made in the calculations.

DODGEVIPER13
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Homework Statement


Given Xe=4.5 and E=.50 MV/m find the displacement flux density

Homework Equations


D=ε0εrE
εr=1+Xe

The Attempt at a Solution


Using the equation above I get (8.85e-12)(5.5)(.50e6)=.000024338 C/m^2 which is not what the answers say they give it in nC which I can't see how they can do?
 
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Sorry they give it in nC/m^2
 
Actually looking at it the real units are μC/m^2
 
Should it be 24.338μC/m^2
 

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