Question

Two charges are located in the x-y plane. If q-4.85 nC and is locatedt0.00 m, y 0.960 m), and the second charge has magnitude of q2 = 3.20 nC and is located at (x = 1.20 m, y = 0.600 m), calculate the x and y components, Ex and E, of the electric field, E, in component form at the origin, (0,0). The Coulomb force constant is 1/(4m) = 8.99 × 109 N·m2/C2 10 Ex=| -1.42949 x1010 N/C Eu-1-1.42949×10 N/C

Answer 1

electric field due to q1

E1 = k q1 /r^2 j

E1 = 9*10^9* 4.85*10^-9 / 0.96^2 j

E1 = 47.363 j

now angle between x axis and E2

theta = arctan ( 0.6/1.2) = 26.565

electric field due to q2

E2 = k q2 / r^2

E2 = 9*10^9*3.2*10^-9 / (1.2^2 + 0.6^2) * ( - cos 26.565 i - sin 26.565 j )

E2 = - 14.3108 i - 7.155 j

net electric field

E = E1 + E2

E = 47.363 j - 14.3108 i - 7.155 j

E = - 14.3108 i + 40.208 j

now

Ex = - 14.3108

Ey = 40.208

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