Electric field of a line of charge with symmetry

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SUMMARY

The discussion centers on calculating the electric field of a line of charge using the equation E=KQ/R², where K is the constant 9x10⁹. The user performed a standard integral of k∫dq/r², incorporating symmetry and recognizing that the electric field's magnitude points in the y-direction. The final expression derived for the electric field is E=√2kλ/a, directed upwards and perpendicular to the line of charge. A critical point raised is the dependency of the radial unit vector, 𝑟̂, on the variable x, which must be accounted for in calculations.

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Familiarity with integral calculus, specifically in the context of physics
  • Knowledge of symmetry in electric field calculations
  • Basic understanding of linear charge density (λ)
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  • Study the derivation of electric fields from continuous charge distributions
  • Learn about the application of the arctangent function in electric field calculations
  • Explore the concept of unit vectors in vector calculus
  • Investigate the effects of charge distribution on electric field directionality
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Students studying electromagnetism, physics educators, and anyone involved in solving problems related to electric fields and charge distributions.

Regtic
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Homework Statement


http://imgur.com/W4Ntkfb

Homework Equations


E=KQ/R2
e= electric field
Q = charge
R = radius from point to charge
K is a constant, 9x109


The Attempt at a Solution


http://imgur.com/4CTEwDw

If my handwriting sucks, I basically did the standard integral of k[itex]\int[/itex]dq /r2 but kept the r hat vector since there is symmetry and I know that the magnitude will be be in the y direction. I used dq=[itex]\lambda[/itex]dx and r2=x2+a2 which gave me an integral that was an arctan integral. By substituting a for x, I got [tex]E= \frac{k \lambda \pi}{2a}[/tex] in the radial direction.

The answer should be √2k[itex]\lambda[/itex]/a upwards and perpendicular to the line of charge
 
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Regtic said:

Homework Statement


http://imgur.com/W4Ntkfb

Homework Equations


E=KQ/R2
e= electric field
Q = charge
R = radius from point to charge
K is a constant, 9x109


The Attempt at a Solution


http://imgur.com/4CTEwDw

If my handwriting sucks, I basically did the standard integral of k[itex]\int[/itex]dq /r2 but kept the r hat vector since there is symmetry and I know that the magnitude will be be in the y direction. I used dq=[itex]\lambda[/itex]dx and r2=x2+a2 which gave me an integral that was an arctan integral. By substituting a for x, I got [tex]E= \frac{k \lambda \pi}{2a}[/tex] in the radial direction.

The answer should be 2k[itex]\lambda[/itex]/R r hat.
##\hat{r}\ ## depends on x, it's not constant.

You failed to take that into account.


Here's the image for the link to your problem.
attachment.php?attachmentid=69770&stc=1&d=1400044878.png


##\hat{r}\ ## depends on x, it's not constant.

You failed to take that into account.
 

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SammyS said:
##\hat{r}\ ## depends on x, it's not constant.

You failed to take that into account.Here's the image for the link to your problem.
attachment.php?attachmentid=69770&stc=1&d=1400044878.png


##\hat{r}\ ## depends on x, it's not constant.

You failed to take that into account.

Oh right it's not constant in direction... was just thinking of that as a constant. Derp, shouldn't be studying this late. Thank you. Redid it with Ey and got it.

also it seems whenever I post photos directly into the forum they become ridiculously large, or maybe that's just photos from Iphone...
 

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