what is the strength of the electric field at the position indicated by the dot?

Let the charge above the x-axis be q1q1 and the charge below be q2q2.

From the diagram, we extract the following quantities,

• Charge q1=+1.0×10−9 Cq1=+1.0×10−9 C located at r=√(0.05 m)2+(0.05 m)2=0.071 mr=(0.05 m)2+(0.05 m)2=0.071 m from the field point.
• Charge q2=+1.0×10−9 Cq2=+1.0×10−9 C located at r=√(0.05 m)2+(−0.05 m)2=0.071 mr=(0.05 m)2+(−0.05 m)2=0.071 m from the field point.

A) Now the net electric field on the point indicated by the dot in the diagram is the vector sum of the two electric fields due to q1q1 and q2q2. Notice that because of symmetry, the vertical components of the electric field cancel out. Now only the horizontal components of the fields add up. Because the point charges and the field point is located in perpendicular axes with the same distance 0.05 m0.05 m, the distance rr between make an angle of α=arctan(0.050.05)=45∘α=arctan⁡(0.050.05)=45∘ towards the axes. Therefore, we can write the electric field as

E=E1cosα+E2cosαE=kq1rcosα+kq1rcosαE=E1cos⁡α+E2cos⁡αE=kq1rcos⁡α+kq1rcos⁡α

Notice that q2=q1q2=q1, therefore

E=2kq1rcosαE=2kq1rcos⁡α

Substituting the quantities,

E=2(8.988×109 N⋅m2/C2)1.0×10−9 C(0.071 m)2cos45∘E≈2.5×10−3 N/CE=2(8.988×109 N⋅m2/C2)1.0×10−9 C(0.071 m)2cos⁡45∘E≈2.5×10−3 N/C

B) From the discussion above, we know that by symmetry only horizontal components of the fields remain. Since both charges are positive, the electric field points towards the positive x-axis or θ=0∘