05 m)
E
−
×
×
=
= 3 6 × 104 N C–1
The electric field vector E2B at B due to the negative charge q2 points
towards the right and has a magnitude
9
2
–2
8
2B
2
(9
10 Nm C
)
(10
C)
(0 15 m)
E
−
×
×
=
=  4 × 103  N C–1
The magnitude of the total electric field at B is
EB = E1B –  E2B = 3 2 × 104 N C–1
EB is directed towards the left