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Calculating magnitude of electric field at center of square

  • #1

Homework Statement



Find the magnitude and direction of net electric field at the center of the square array of charges. Find [itex] E_x and E_y [/itex]

The square array of charges http://postimg.org/image/4gf94ymmf/

The Attempt at a Solution


[/B]
My attempt at drawing in the force vectors http://postimg.org/image/mae0fm1d9/ . Now the +3q and +q's should contribute a net of zero to the electric field and they can be ignored. So, we have [itex] E = k [ -2q^2 / d^2 + q^2/d^2 + 5q^2/d^2] = 4k q^2/d^2[/itex]

Taking the x-component of the field we have
[itex] E_x = (4k q^2/d^2)cos(45)[/itex]

the y-component:
[itex] E_x = (4k q^2/d^2)sin(45)[/itex]

This isn't right or I would not be posting here... so what am I doing wrong? I'm really not understanding how to do these types of problems for net fields in squares. Have I even drawn the vectors correctly?
 

Answers and Replies

  • #2
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  • #3
The missing term is the -5q charge, yes? So it should be
[itex]E=k[−2q/(\sqrt2 * d)+q/(\sqrt 2 * d) + 2(5q/(\sqrt 2 * d))= k[9q / (\sqrt 2 * d)][/itex]

Also, changed q^2 to q since this is the electric field not force... silly mistake, and the bottom term should be [itex] \sqrt 2 * d [/itex] instead of d^2?

This is still incorrect is not? I'm really not seeing what I'm missing here.
 
  • #4
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Methinks that should do almost do it (dumped the picture, so I'm going by memory); the "two" and "one" are both on axes? In which case "d."
 
  • #5
Methinks that should do almost do it (dumped the picture, so I'm going by memory); the "two" and "one" are both on axes? In which case "d."
Ah fair point forgot about those being strictly on the x-axis, so need to fix that then resolve into E_x and E_y for the respective terms and then take the square root of those squared for the magnitude.
 

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