Why is there a minus sign in the vector sum of two perpendicular vectors?

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Homework Help Overview

The discussion revolves around the vector sum of two perpendicular vectors, specifically \(\vec{E}_r\) and \(\vec{E}_{\theta}\), and the confusion regarding the presence of a minus sign in the equation for their resultant vector \(\vec{E}_n\).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the correct formulation of the vector sum, questioning the use of a plus versus a minus sign in the equation. There is an attempt to clarify the geometric interpretation of the vectors involved, with references to right triangles and the Pythagorean theorem.

Discussion Status

The discussion is ongoing, with multiple interpretations being explored regarding the direction of the vectors and the implications of the minus sign. Some participants are seeking clarification on the assumptions made about the orientation of the vectors.

Contextual Notes

There is mention of an attachment that is pending approval, which may contain relevant visual information. Participants are also noting the potential for differing conventions in vector representation, particularly regarding the direction of \(\vec{E}_{\theta}\).

_Andreas
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Homework Statement



The vector [tex]\vec{E}_n[/tex] is the vector sum of the two vectors [tex]\vec{E}_r[/tex] and [tex]\vec{E}_{\theta}[/tex], which are perpendicular to each other (see attached picture). Calculate the magnitude of [tex]\vec{E}_n[/tex].

The Attempt at a Solution



[tex]E_n=E_r\cos(\theta)+E_{\theta}\sin(\theta)[/tex]

But this is wrong. It's supposed to be a minus sign instead of a plus. I have no idea why. I thought it would be to simply add the components of [tex]\vec{E}_r[/tex] and [tex]\vec{E}_{\theta}[/tex] along the direction of [tex]\vec{E}_n[/tex]. Why the minus sign?
 

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_Andreas said:

Homework Statement



The vector [tex]\vec{E}_n[/tex] is the vector sum of the two vectors [tex]\vec{E}_r[/tex] and [tex]\vec{E}_{\theta}[/tex], which are perpendicular to each other (see attached picture). Calculate the magnitude of [tex]\vec{E}_n[/tex].

The Attempt at a Solution



[tex]E_n=E_r\cos(\theta)+E_{\theta}\sin(\theta)[/tex]
The attachment is still pending approval, so I can't see the image you attached. Based on your description, I have a right triangle with E_n as the hypotenuse and E_r and E_theta as the legs of the triangle. E_r is the base of the triangle and E_theta is the altitude.

If my interpretation of your description is correct, your equation just above is incorrect. You've omitted the vector "caps" in the equation above, so I don't know if E_n is now supposed to be a magnitude.

In any case, you have E_n = E_r + E_theta (vectors).
So |E_r| = |E_r| cos theta and |E_theta| = |E_r| sin theta.
|E_n| = sqrt{|E_r|^2 + |E_theta|^2}
_Andreas said:
But this is wrong. It's supposed to be a minus sign instead of a plus. I have no idea why. I thought it would be to simply add the components of [tex]\vec{E}_r[/tex] and [tex]\vec{E}_{\theta}[/tex] along the direction of [tex]\vec{E}_n[/tex]. Why the minus sign?
Until I can see the drawing, I don't know.
 
_Andreas said:
[tex]E_n=E_r\cos(\theta)+E_{\theta}\sin(\theta)[/tex]
You are assuming that [tex]\vec{E}_n[/tex] is horizontal. If true, then your equation is OK. In general, use the Pythagorean theorem to find the magnitude as Mark44 stated.

But this is wrong. It's supposed to be a minus sign instead of a plus. I have no idea why. I thought it would be to simply add the components of [tex]\vec{E}_r[/tex] and [tex]\vec{E}_{\theta}[/tex] along the direction of [tex]\vec{E}_n[/tex]. Why the minus sign?
Could it be that you have the direction of [tex]\vec{E}_{\theta}[/tex] reversed?
 
Doc Al said:
You are assuming that [tex]\vec{E}_n[/tex] is horizontal. If true, then your equation is OK. In general, use the Pythagorean theorem to find the magnitude as Mark44 stated.

Yes, [tex]\vec{E}_n[/tex] is horizontal. But about the magnitude: the result seems to be a scalar (negative surface charge density divided by permittivity).

Doc Al said:
Could it be that you have the direction of [tex]\vec{E}_{\theta}[/tex] reversed?

I've drawn the picture exactly as it appears in my book, so I doubt that. However, it does indeed seem to be standard practice to draw it in the opposite direction (opposite to the direction in which it is drawn in my book, that is).
 

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