- #1

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**What the hell is wrong with this?! :)**

I just can't spot the mistake! Please see if you can.

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- Thread starter LucasGB
- Start date

- #1

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I just can't spot the mistake! Please see if you can.

- #2

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- #3

- 649

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In your first equation, both the dz parts have the wrong sign. Remember, you are going anti-clockwise, so the first dz-part moves in the negative z-direction, so it should be negative, not positive. The same (or rather, opposite) goes for the other dz-part at x=0.

EDIT: Don't be fooled by the way the arrows are drawed. If E_z is positive in the second term, then it points backward in comparison to the direction of the path, and will therefore contribute negatively, hence the correct expression is -E_z*dz

Torquil

- #4

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EDIT: Don't be fooled by the way the arrows are drawed. If E_z is positive in the second term, then it points backward in comparison to the direction of the path, and will therefore contribute negatively, hence the correct expression is -E_z*dz

Torquil

But I don't think the signs are wrong. It follows from the properties of dot products. Please check the attachment.

- #5

Ich

Science Advisor

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Sorry, LucasGB, this may sound stupid, but yes, please check the attachment.Please check the attachment.

- #6

- #7

- 649

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No I'm pretty sure the signs are wrong. In your second calculation the error manifests itself like this:

Notice how you have defined your angles. Your first and second equation is correct. But when you write it in terms of the components E_x and E_z you are using the wrong sign for both the E_z's.

E.g. the second term: You have defined the theta angle to be the angle between E_2 and -dz. Therefore |E_2|cos(theta) is -E_{2z}, not E_{2z}. The opposite goes for the other E_z part.

You get the component E_z of a vector E by doing |E|cos(theta) where theta is the angle between the vector and the "positively pointing" z-axis. Your are assuming the opposite for the two parts that are related to z-components.

Torquil

- #8

- 181

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That's perfect. I can totally see it now. Thank you very much, torquil, you're the man!

I just learned something new!

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