13 × 106  N C–1  along BP
Field at P due to charge –10 mC
–5
12
2
1
2
10
C
4 (8 854 10
C N
m
)
−
−
−
=
π
×
 
2
4
2
1
(15
0 25)
10
m
−
×
+
×
=  3 86 × 106  N C–1 along PA
The resultant electric field at P due to the two charges at A and B is
=  2