No Limits of Integration for Electric Field Integral?

In summary: No, it does not have a width because the calculation assumes that it is a "thin" ring which means zero width.
  • #1
ChiralSuperfields
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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,

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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!
 
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  • #2
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?
 
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  • #3
kuruman said:
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
kuruman said:
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
Callumnc1 said:
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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Related to No Limits of Integration for Electric Field Integral?

1. What is the concept of "No Limits of Integration for Electric Field Integral" in physics?

The concept of "No Limits of Integration for Electric Field Integral" refers to the fact that the electric field integral can be calculated over any arbitrary closed surface, regardless of its shape or size. This means that there are no limits to the integration in terms of the shape or size of the surface, as long as it is closed.

2. How is the electric field integral related to Gauss's Law?

The electric field integral is a mathematical representation of Gauss's Law, which states that the electric flux through any closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. This relationship allows us to calculate the electric field at a point by integrating over a closed surface surrounding that point.

3. Can the electric field integral be used to calculate the electric field inside a conductor?

Yes, the electric field integral can be used to calculate the electric field inside a conductor, as long as the surface used for integration is chosen to enclose the entire volume of the conductor. This is because the electric field inside a conductor is zero, and therefore the electric flux through any closed surface surrounding it is also zero.

4. Are there any limitations to using the electric field integral for calculating electric fields?

One limitation of using the electric field integral is that it assumes a static electric field. This means that it cannot be used to calculate the electric field in situations where the charges are moving or changing over time. Additionally, it is only applicable in situations where Gauss's Law holds true, such as in the absence of magnetic fields.

5. How is the electric field integral used in practical applications?

The electric field integral is used in many practical applications, such as in the design of electronic circuits, the calculation of electric potential in electrostatics, and the analysis of electric fields in materials. It is also an important tool in understanding the behavior of lightning and other atmospheric electric phenomena.

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