05m)
E
−
×
×
=
 =  3 6 × 104  N C–1
The electric field vector E2A at A due to the negative charge q2 points
towards the right and has the same magnitude Hence the magnitude
of the total electric field EA at A is
EA = E1A + E2A = 7 2 × 104 N C–1
EA is directed toward the right