(b) Field at Q due to charge + 10 mC at B
=
5
12
2
1
2
10
C
4
(8 854
10
C
N
m
)
−
−
−
−
π
×
2
2
4
2
1
[15
(0 25) ]
10
m
−
+
×
×
=  3 99 × 106  N C–1 along BQ
Field at Q due to charge –10 mC at A
=
5
12
2
1
2
10
C
4
(8