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

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)