No Limits of Integration for Electric Field Integral?

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  • #1
Callumnc1
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Homework Statement:
This is a more concise version of my 'Electric Field of a Uniform Ring of Charge' thread that I posted yesterday and made a typo.
Relevant Equations:
Continuous charge distribution formula
For this problem,

s7y2w.png
uCKQS.png


The solution is,

ok160.png


However, why have they not included limits of integration? I think this is because all the small charge elements dq across the ring add up to Q.

However, how would you solve this problem with limits of integration?

Many thanks!
 

Answers and Replies

  • #2
kuruman
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You can assume a uniform linear charge density on the ring, ##\lambda=\dfrac{Q}{2\pi a}.## Then an arc element ##ds## at polar angle ##\phi## in the plane of the ring subtends angle ##d\phi## so that ##ds=a d\phi##. The charge on that element is ##dq=\lambda ds=\lambda a d\phi## so that when you integrate going around the ring once, you have $$\int dq=\int_0^{2\pi}\lambda a d\phi=2\pi a\lambda=2\pi a\frac{Q}{2\pi a}=Q$$Does this answer your question?
 
  • #3
Callumnc1
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You can assume a uniform linear charge density on the ring, ##\lambda=\dfrac{Q}{2\pi a}.## Then an arc element ##ds## at polar angle ##\phi## in the plane of the ring subtends angle ##d\phi## so that ##ds=a d\phi##. The charge on that element is ##dq=\lambda ds=\lambda a d\phi## so that when you integrate going around the ring once, you have $$\int dq=\int_0^{2\pi}\lambda a d\phi=2\pi a\lambda=2\pi a\frac{Q}{2\pi a}=Q$$Does this answer your question?
Thanks for your answer @kuruman, I think that answers my question for now!
 
  • #4
Callumnc1
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You can assume a uniform linear charge density on the ring, ##\lambda=\dfrac{Q}{2\pi a}.## Then an arc element ##ds## at polar angle ##\phi## in the plane of the ring subtends angle ##d\phi## so that ##ds=a d\phi##. The charge on that element is ##dq=\lambda ds=\lambda a d\phi## so that when you integrate going around the ring once, you have $$\int dq=\int_0^{2\pi}\lambda a d\phi=2\pi a\lambda=2\pi a\frac{Q}{2\pi a}=Q$$Does this answer your question?
Sorry @kuruman, why are you allowed to use the linear charge density, because doesn't the ring have a width too?

Are you assuming that x >> width, so the width is negligible?

Many thanks!
 
  • #5
kuruman
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Sorry @kuruman, why are you allowed to use the linear charge density, because doesn't the ring have a width too?
No, it does not have a width because the calculation assumes that it is a "thin" ring which means zero width. If it has a width, say it is a washer of inner radius ##a## and outer radius ##b##, then you have to do a different calculation and also integrate over ##r## from ##a## to ##b##. That's a different homework problem.
 
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