Magnitude and DIRECTION of electric field of a square.

  • Thread starter johnknee
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Homework Statement


Calculate the magnitude of the electric field at the center of a square 42.5cm on a side if one corner is occupied by a −38.2μC charge and the other three are occupied by −26.9μC charges. (This part done)

Choose the correct direction of the electric field at the center of the square.
A.outward of −38.2μC charge
B.perpendicular to the diagonal passing through −38.2μC charge
C.toward to −38.2μC charge

Not sure how to approach the directions part.

Homework Equations


E = kQ/r^2
Magnitude E = ((Ex)^2 + (Ey)^2)^1/2


The Attempt at a Solution


r = 0.425m
E1 = kQ1/r^2 = ((9*10^9)(38.2*10^-6))/0.3^2 = 3.8*10^6 N/C
E2 = E3 = E4 = ((9*10^9)(26.9*10^-6))/0.3^2 = 2.7*10^6 N/C

I then solved each of their x and y components, and added them together. Net Ex is -787000 and net Ey is 787000.

Resultant = ((-787000)^2 + (787000)^2)^1/2 = 1.11*10^-6 N/C

I am not sure how I would figure out the direction part.
*I am thinking towards the - 38.2 charge
 
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Answers and Replies

  • #2
haruspex
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I am thinking towards the - 38.2 charge
The direction of the field is the direction a positive charge would tend to move if placed in the field. Does that fit with your answer? (I'm not hinting either way.)
 

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