### What is the direction of the electric field at the position indicated by the dot in (Figure 1)?

### How can we determine the field’s direction at the location shown by the dot (Figure 1.)?

**Answer:**

Because the charges are located in the same distance and both positive, they repel at the same rate. If you draw the forces they will cancel the part that is y in the field that is final. It will only be the component x. Since they are identical, it will result in twice as large as the component x in one electric field.

E=kq/r^2

R= 5 x root2 equals 7.07 cm , or .0707 millimeters (property of right triangle 45-45-90)

E=(9 x 10^9)(1 x 10^-9 )/(.0707)^2

= 1800.5 N/C

To find the x component

cos 45 = x/1800.5

x= 1273.18 N/C

Simply multiply the number by 2 to accommodate each charge,

E=2546.35 N/C with a direction of 0 ° (no y component because they cancel one another out)