Determine the magnitude and direction of the electric field intensity at Point p in the diagram

Question

            . . . . . . . . . . . p

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  6cm. . . . . . . . . . . . | . . . . . . . . . . . .   6cm 

B________________________________________A

(+14nC) . . . . . . . . . . . . . . . . . . . . . . .(+12nC) 

Answer 1

I suggest finding the x and y components of the field from each charge, then finding the x and y components of the resultant field vector, and then getting the magnitude and direction from them.

 

Field at P due to charge at A:

 

E_Ax = (kq_A/r_A²)cosθ_A

 

q_A = 14 x 10^-9 C

r_A = √(6² + 8²) cm = 10 cm = 0.10 m ⇒ r_A² = 0.01 m²

cosθ_A = 6/10 = 0.6

kq_A/r_A² = (9 x 10⁹)(14 x 10^-9) / (0.01) N/C = 12600 N/C

E_Ax = (kq_A/r_A²)cosθ_A = 12600*0.6 N/C = 7560 N/C

sinθ_A = 8/10 = 0.8

E_Ay = (kq_A/r_A²)sinθ_A = 12600*0.8 N/C = 10080 N/C

 

Field at P due to charge at B:

 

E_Bx = -(kq_B/r_B²)cosθ_B

q_B = 12 x 10^-9 C

r_B² = r_A² = 0.01 m²

cosθ_B = 6/10 = 0.6
kq_B/r_B² = (9 x 10⁹)(12 x 10^-9) / (0.01) N/C = 10800 N/C
E_Bx = -(kq_B/r_B²)cosθ_B = -10800*0.6 N/C = -6480 N/C
sinθ_B = 8/10 = 0.8
E_By = (kq_B/r_B²)sinθ_B = 10800*0.8 N/C = 8640 N/C

 

Resultant field at P:

 

E_Px = E_Ax + E_Bx = (7560 - 6480) N/C = 1080 N/C

E_Py = E_Ay + E_By = (10080 + 8640) N/C = 18720 N/C

 

E_p = √(E_Px² + E_Py²) = √(1080² + 18720²) N/C ≅ 18751 N/C

 

θ_P = tan^-1(E_Py / E_Px) = tan^-1(18720 / 1080) ≅ 86.7° above the positive x-axis