Potential Difference Problem - setting up the integral

In summary, the conversation is about a problem involving the equation V=kq/x. The person had made a mistake in their initial attempt at solving the problem by using a different integration variable, causing their answer to be incorrect. Another person explains that this is because the variable was changed and that they can avoid this mistake by using a different variable name. The conversation ends with gratitude for the explanation.
  • #1
Taulant Sholla
96
5

Homework Statement


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Homework Equations


V=kq/x​

The Attempt at a Solution


I know the correct solution. It's...​
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On my first attempt, rather than use (d+x) in the denominator and integrate from 0 to L, I instead used (x) and integrated from (d) to (L+d).
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This produces the wrong answer, but why does it produce the wrong answer? Both approaches seem to capture the setup and handle the integration correctly. Obviously I'm wrong, but - again - why/how am I wrong? Thank you.
 

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  • #2
##\lambda## is no longer ##cdx## if you change variables. You can avoid this kind of oversight by using a different integration variable name, e.g. ##u ## or ##v##, but not the same name ##x##
 
  • #3
Thank you! However, question: where did I "change variables" and/or why is λ no longer cdx?
 
  • #4
You changed the integration variable ##x## that runs from 0 to L into a variable with name e.g. ##u## that runs from d to d + L. In other words ##u = x+d##. When ##x = 0##, ##\lambda = cx=0##, but when ##u=d##, then ##\lambda \ne cu##.
 
  • #5
Ah, thank you so much. This is very helpful. I appreciate your explanation very much!
 
  • #6
You're welcome. That's what PF is for... :biggrin:
 

Related to Potential Difference Problem - setting up the integral

1. What is a potential difference problem?

A potential difference problem is a type of problem in physics and engineering that involves calculating the potential difference (also known as voltage) between two points in an electric circuit. It is an important concept in understanding and analyzing electrical circuits.

2. How do you set up the integral for a potential difference problem?

To set up the integral for a potential difference problem, you first need to identify the points between which the potential difference is being measured. Then, you need to determine the path along which the integral will be calculated. This path is typically drawn as a line connecting the two points. Finally, you need to determine the function that describes the electric field along this path, which will be used in the integral calculation.

3. What is the equation for calculating potential difference?

The equation for calculating potential difference is V = W/Q, where V is the potential difference, W is the work done in moving a charge Q from one point to another, and Q is the charge. This equation can be used for both direct and alternating current circuits.

4. Can potential difference be negative?

Yes, potential difference can be negative. This occurs when the direction of current flow is opposite to the direction of the electric field. In this case, the potential difference is said to be negative because the work done in moving the charge is in the opposite direction of the electric field.

5. How is potential difference related to electric potential energy?

Potential difference and electric potential energy are closely related. Potential difference is a measure of the change in electric potential energy per unit charge between two points in an electric field. In other words, it represents the amount of work required to move a unit charge from one point to another. This relationship is described by the equation V = ΔU/Q, where V is the potential difference, ΔU is the change in electric potential energy, and Q is the charge.

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