Clarification on an electric fields solution

In summary, the conversation is about a question regarding a solution provided in a solutions manual for a homework problem. The question is about why the radius "a" loses its exponent when converting dE into cylindrical coordinates. The answer is that it comes from the arc length formula, d\ell=a\cdot d\theta.
  • #1
Allenman
58
0
This isn't actually a homework problem. I just had a question about the solution they provided.

Homework Statement


physprob96.png



2. Solution given in solutions manual
physsol96.png



3. My question

When they convert dE into cylindrical coordinates why does the radius "a" lose its exponent?

dE = [itex]\frac{\kappa\lambda\delta l}{a^{2}}[/itex] = [itex]\frac{\kappa\lambda\delta\theta}{a}[/itex]

I have to be missing something simple, I just know it...
 
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  • #2
Allenman said:
This isn't actually a homework problem. I just had a question about the solution they provided.

Homework Statement


physprob96.png


2. Solution given in solutions manual
physsol96.png


3. My question

When they convert dE into cylindrical coordinates why does the radius "a" lose its exponent?

dE = [itex]\frac{\kappa\lambda\delta l}{a^{2}}[/itex] = [itex]\frac{\kappa\lambda\delta\theta}{a}[/itex]

I have to be missing something simple, I just know it...
It's because [itex]d\ell=a\cdot d\theta\,.[/itex]
 
  • #3
Does that come from the arc length formula?

Thank you
 
  • #4
Yes it does.
 
  • #5


As a scientist, it is important to carefully analyze and understand the solutions provided in textbooks or manuals. In this case, it seems that the solution manual has made a mistake in converting the electric field from Cartesian to cylindrical coordinates. The correct expression for the electric field in cylindrical coordinates should be dE = \frac{\kappa\lambda\delta l}{a} instead of dE = \frac{\kappa\lambda\delta l}{a^{2}}. This is because in cylindrical coordinates, the distance from the source is given by the radius "a" and not the squared radius as in Cartesian coordinates. This may be a simple mistake, but it is important to double check solutions and understand the reasoning behind the conversions to ensure accuracy in our scientific calculations.
 

1. What is an electric field solution?

An electric field solution is a mathematical representation of the electric field in a given region. It is a vector field that describes the strength and direction of the electric force on a charged particle at any point in space.

2. How is an electric field solution calculated?

An electric field solution is calculated using Coulomb's law, which states that the electric force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

3. What are some applications of electric field solutions?

Electric field solutions have many practical applications, including in the design and optimization of electronic devices, in medical imaging and treatment, and in the study of atmospheric phenomena such as lightning and auroras.

4. How does an electric field solution differ from an electric potential solution?

An electric field solution describes the strength and direction of the electric field at every point in space, while an electric potential solution describes the potential energy per unit charge at a given point in space. The two are related, as the electric potential is the negative gradient of the electric field.

5. Can an electric field solution exist without a source charge?

No, an electric field solution cannot exist without a source charge. The electric field is created by charged particles, so without a source charge there would be no electric field. However, there can be regions of space where the electric field is zero, known as electric field lines, even in the absence of a source charge.

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