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En = k [Qn / Dn^2] [ cos Tn i + sin Tn j ]
D1 = sqrt(4^2 + 3*2) = 5
D2 = 5
D3 = 4
cos T1 = 4/5, sin T1 = 3/5
cos T2 = 0, sin T2 = oo (force straight up)
cos T3 = 4/4 = 1 , sin T2 = 0 (force horizontal)
plug and chug, add the x and y components of En up to get total E at Q4
for force multiply Each En by Qn and add components again
Find (a) the electric field, direction and magnitude, at P (4.00 m, 5.00m) in Fig. 4 below due to q1 = +5.00 x 10-6 C at (0, 2.00 m), q2 = -3.00 x 10-6 C at (4.00 m, 0), and q3 = +1.6 x 10-6 C at (0, 5.00 m) and (b) the force on q4 = +2.0 x 10-6 C placed at P.
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