Physics question about electric field and vectors.?

Two equal charges of 5 C are located in the xy-plane, one at (0m, 57.9) and the other at (75.9m, 0m).


a) find the magnitude of the electric field at the origin to these two charges.


b) find the angle between the vector for the electric field and the origin and a vector in the minus x-direction. Answer in degrees.





I found the answer to part a) to be - 1.8179E-5 by using coulombs law and then the Pythagorean theorem.





i tried -tan for the angle and got 30 degrees but thats incorrect.





please if u know how to do this problem explain it to me.. i need to this hw by tonight and i cant find it anywhere else.





thanks in advance

Physics question about electric field and vectors.?
let coordinates of point charges be (x,0) %26amp; (0,y), and let unit vectors along (+x,+y)) be (+i, +j)


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placing a +ve charge (qo) at origin.


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(x,0) charge will tend to repel test charge along -x direction with force (Fx) linked to electric vector component (Ex) by


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Fx = qo Ex = 9*10^9 *5qo/x^2


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(0,y) charge will tend to repel test charge along -y direction with force (Fy) linked to electric vector component (Ey) by


Fy = qo Ey = 9*10^9 *5qo/y^2


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net electric field vector


E = Ex (-i) + Ey (-j)


E = - 9*10^9 *5 [ i /x^2 + j /y^2]


put values of x, y


E = - 9*10^9 *5 [ i /75.9^2 + j /57.9^2]


E = - 10^6[ 45000 i /5760.81 + 45000 j /3352.41]


E = 10^6[ - 7.81 i - 13.42 j] %26gt;%26gt;%26gt;%26gt;%26gt;%26gt;


magnitude of (E)


|E| = 10^6[7.81^2 + 13.42^2]^1/2 = 15.53*10^6 N/C


direction


tan (p) = Ey/Ex = - 13.42*10^6/ - 7.81*10^6 = 1.7183


p = angle made by (E) vector with (-x) axis measued anticlockwise = tan^-1[1.7183] = 59.8 degree

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