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What is the electric field at the point?

  • Thread starter Ammora
  • Start date
  • #1
17
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Homework Statement



Two tiny objects with equal charges of 7.48 µC are placed at the two lower corners of a square with sides of 0.350 m, as shown.

Image: http://www.webassign.net/grrphys2/16-p-039.gif

Find the electric field at point A, the upper left corner. (Assume the positive x-direction points to the right.)
magnitude: ____N/C
direction: ____° CCW from the positive x-axis

Homework Equations



kq/r^2

The Attempt at a Solution



I already got the magnitude right, here's what I did:
kq/r^2
= 8.99e9 (7.48e-6)/(.350)^2 = 5.49e+5

Where I'm stuck is the direction.
I tried tan theta = (.350) and that was wrong, and I tried tan theta= (.35/2)/.35 and that was wrong also.
Any help would be greatly appreciated. This is due tonight. Thanks in advance.
 

Answers and Replies

  • #2
gneill
Mentor
20,793
2,773
I already got the magnitude right, here's what I did:
kq/r^2
= 8.99e9 (7.48e-6)/(.350)^2 = 5.49e+5
That only accounts for the field due to the charge in the lower left corner. What about the other charge?

Also, because you're dealing with a square with sides parallel to the coordinate axes, you should be able to use ratios of side lengths (similar triangles) in lieu of trig functions to work out the force components, if you wish.
 

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