Continuous line charge distribution

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SUMMARY

The discussion centers on calculating the electric field at a point P due to a continuous line charge distribution from a rod of length 80 cm with a uniform linear charge density of 5 mC/m. The user attempts to derive the electric field using the formula E = k * linecharge(dL) * (h/r) / r^2, integrating over half the rod due to symmetry. However, discrepancies arise in the final calculations, with one participant suggesting that the integration may have been performed incorrectly, resulting in an answer that is double the expected value.

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Familiarity with calculus, particularly integration techniques
  • Knowledge of linear charge density concepts
  • Basic physics principles related to symmetry in charge distributions
NEXT STEPS
  • Review integration techniques for electric field calculations
  • Study the derivation of electric fields from continuous charge distributions
  • Explore the concept of symmetry in electrostatics
  • Learn about the application of Coulomb's law in complex charge configurations
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Students of physics, electrical engineers, and anyone involved in electrostatics or electric field calculations will benefit from this discussion.

GravityGirl
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A rod of length 80 cm has a uniform linear charge density of 5 mC/m. Determine the Electric Field at a point P located at a perpendicular distance 57 cm along a line of symmetry of the rod


i don't know what i wrong..but here is what i am doing

linecharge(change in lenght)= change in Q

E=Kdq/r^2
E=k*linecharge(dL)/r^2

i drew a free body diagram and the x components cancle out so i only have to worry about y

Ey=k*linecharge(dL)(costhea)/r^2 *costheta=height in y direction/r
Ey=k*linecharge(dL)(h/r)/r^2

so now i am going to integrate only half of the rod because of symetry

so i have the integral from 0 to L/2 of k*linecharge(dL)(h/r)/r^2

i simplify and i have Ey=k*linecharge(h) integral from 0 to L/2 of 1/r^3

i simplify again to get =k*linecharge(L/2)/h(h^2+(L/2)^2)^1/2

i believe this is the correct formula...but when i put the numbers in i get the wrong answer

please help
 
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Hello,

I obtain an answer that is double of yours.
Please check your integration carefully.


Best regards
 
variation said:
Hello,

I obtain an answer that is double of yours.
Please check your integration carefully.


Best regards

I get something completely different.
 

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