Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Vector sum, polar coordinates

  1. Jan 29, 2009 #1
    1. The problem statement, all variables and given/known data

    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].

    3. The attempt at a solution


    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?

    Attached Files:

  2. jcsd
  3. Jan 30, 2009 #2


    Staff: Mentor

    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}
    Until I can see the drawing, I don't know.
  4. Jan 30, 2009 #3

    Doc Al

    User Avatar

    Staff: Mentor

    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.

    Could it be that you have the direction of [tex]\vec{E}_{\theta}[/tex] reversed?
  5. Jan 30, 2009 #4
    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).

    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).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook