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Pleas help electric field

  • #1
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pleas help!! electric field

a semi infinite insulting rod carries a constant charge per unit length of lambda. SHow that he electric field at the point P makes an angle of 45 with the rod and that this result is independant of R

[tex] q = \lambda x [/tex]
[tex] dq = \lambda dx [/tex]

for the Y direction poiting downward
[tex] dE_{y} = \int_{0}^{\infty} \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{R^2 + x^2} \sin{\theta} [/tex]
[tex] \sin{\theta} = \frac{x}{R^2 + x^2} [/tex]
[tex] dE_{y} = \int_{0}^{\infty} \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda R dx}{(R^2 + x^2)^\frac{3}{2}} [/tex]
[tex] dE_{y} = \frac{\lambda R}{4 \pi \epsilon_{0}} \int_{0}^{\infty} \frac{dx}{(R^2 + x^2)^\frac{3}{2}} [/tex]
[tex] dE_{y} = \frac{\lambda R}{4 \pi \epsilon_{0}} [\frac{x}{R^2 \sqrt{R^2+x^2}}]_{0}^{\infty} [/tex]
now im not sure about this integral and whether or not it converges...

for the horizontal pointing left
[tex] \frac{\lambda R}{4 \pi \epsilon_{0}} \int_{0}^{\infty} \frac{xdx}{(R^2 + x^2)^\frac{3}{2}} [/tex]
same problem here

please help!
 

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Answers and Replies

  • #2
quasar987
Science Advisor
Homework Helper
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stunner5000pt said:
a semi infinite insulting rod carries a constant charge per unit length of lambda. SHow that he electric field at the point P makes an angle of 45 with the rod and that this result is independant of R

[tex] q = \lambda x [/tex]
[tex] dq = \lambda dx [/tex]

for the Y direction poiting downward
[tex] dE_{y} = \int_{0}^{\infty} \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{R^2 + x^2} \sin{\theta} [/tex]
[tex] \sin{\theta} = \frac{x}{R^2 + x^2} [/tex]
[tex] dE_{y} = \int_{0}^{\infty} \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda R dx}{(R^2 + x^2)^\frac{3}{2}} [/tex]
[tex] dE_{y} = \frac{\lambda R}{4 \pi \epsilon_{0}} \int_{0}^{\infty} \frac{dx}{(R^2 + x^2)^\frac{3}{2}} [/tex]
[tex] dE_{y} = \frac{\lambda R}{4 \pi \epsilon_{0}} [\frac{x}{R^2 \sqrt{R^2+x^2}}]_{0}^{\infty} [/tex]
now im not sure about this integral and whether or not it converges...
The integral is right and the limit to evaluate is almost trivial: pull x² out of the square root.
 
  • #3
1,444
2
ok for the Y part
[tex] E_{y} = \frac{\lambda}{4 \pi \epsilon_{0}R} [/tex]
i swhat i got the Y part. Am i right??

for the X part
i got the same thing.
However the resultant for these two given a 45 degree angle... however it isn ot independat of R as in
[tex] E = \sqrt{2} \frac{\lambda}{4 \pi \epsilon_{0}R} [/tex]
 

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