25) ] 10
m
−
+
×
×
=  3 99 × 106  N C–1 along QA Clearly, the components of these two forces with equal magnitudes
cancel along the direction OQ but add up along the direction parallel
to BA Therefore, the resultant electric field at Q due to the two
charges at A and B is
= 2 × 
6
–1
2
2
0