What is the strength of the electric field at the position indicated by the dot in the figure?

electric-field

#1

What is the strength of the electric field at the position indicated by the dot in the figure?
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What is the strength of the electric field at the position indicated by the dot in the figure?
E = _____ N/C.

What is the direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above the horizontal line.
θ = ____ ⁰.

Answer:

Since the charges are at equal distances and both are positive, they will repel with the same magnitude. If you draw the forces, then they will cancel out the y component of the final field. There will only be the x component. And since they are the same, then the final electric field will be twice the x component of one electric field of a charge.
E=kq/r^2
r= 5 x root2 = 7.07 cm or .0707 m (property of 45-45-90 right triangle)
E=(9 x 10^9)(1 x 10^-9)/(.0707)^2
= 1800.5 N/C
Now to find the x component,
cos 45 = x/1800.5
x= 1273.18 N/C
Now just multiply by 2 to accomodate both charges,
E=2546.35 N/C in direction of 0 degrees (no y component since they cancel each other out)