Question

(a) Place a charge of -5.30

Answer 1

You have a right triangle
(because 3^2 + 4^2 = 5^2) , with
q1 at the right angle and
q2 at the end of the shorter leg.
E1(due to q1) = k (7.25 x 10 ^(-6) C) / (0.3 m)^2 = 724 kN/C
EP(due to P ) = k (-8.7 x 10 ^(-6) C) / (0.5 m)^2 = -312 kN/C
Here k = 8.987 x 10^9 N m^2 /C^2

If you position q1 at the origin,
and q2 at (0.3 m, 0)
and P at (0, 0.4 m),
Then E1(due to q1) is directed in the +x direction,
but EP(due to charge at P) must be multiplied by a
unit vector in the direction ( -3i/5 + 4j/5)

So the magnitude of the field at q1 is

1 kN/C times sqrt { [724 - (3/5)312]^2 + [(4/5) 312]^2 } = 592 kN/C