Is My Calculation of Work Along a Defined Curve Correct?

In summary, the article discusses the process of calculating work done along a defined curve in physics, particularly in the context of vector fields. It emphasizes the importance of correctly parameterizing the curve and applying the work integral formula. The author provides examples to illustrate common pitfalls and emphasizes the need for careful evaluation of both the path and the force vector involved in the work calculation to ensure accurate results.
  • #1
Morle
2
0
Homework Statement
I would like to confirm whether my solution is correct.
Relevant Equations
vector line integral
I created a problem for myself, which I have documented in LaTeX. I hope it's acceptable that I included snapshots of my pages.

I would like to calculate the Work W along curve C1.
1727268611164.png


I have solved the problem as shown below. Is my solution correct (see Equation 1.11)?
1727268640213.png


The pictures I included are of low quality, so I've attached the main.pdf for your reference.
 

Attachments

  • main.pdf
    195.8 KB · Views: 15
Physics news on Phys.org
  • #2
I am not opening unsolicited pdfs as a matter of principle.

Please note that the forum has LaTeX compatibility. You’ll have to change a couple of delimiters, but displaying equations written in LaTeX generally works quite well.
 
  • #3
Your final result seems to have the wrong units.
But what's the point of doing this?
 
  • #4
Hello Nasu, The goal is to improve my understanding by practicing vector line integral problems in physics, specifically those related to electrostatics. Why is the Unit of W not correct? q in [As], Q in [As], e0 in [As]/[Vm] so k in [As]*[Vm] and finally W in [As]*[Vm]/[m]=[VAs]=[Ws]=[Nm]=[J]. So the unit is correct, right?
Thanks and best regards.
 
  • #5
What is the meaning of "m" in The denominator of your final result? It looks like "k" divided by some quantity in meters. Which does not have dimensions of work.
 

Similar threads

Back
Top