Question

9. Two point-like charges are placed at the corners of a square as shown in the figure, a-56.0 cm. Find the magnitude of the electric field at the bottom left corner. Let qi = +1 .90 μC, and q-= +2.70 pC N/C

Answer 1

The diagram of the charges is as follows. From the diagram, the distance is, r = squareroot (0.56 m)^2 + (0.56 m)^2 = 0.792 m The angle is, theta = tan^-1 (a/a) = tan^-1(0.56/0.56) = 45 degree The x component of the net electric field is, E_x = -k q_1/r^2 cos 45 degree + kq_2/a^2 = -(9 times 10^9) (1.90 times 10^-6 C)/(0.792 m)^2 cos 45 degree + (9 times 10^9) (2.70 times 10^-6 C)/(0.56 m)^2 = -19.28 times 10^3 + 77.79 times 10^3 = 58.51 times 10^3 i^hat The y component of the net electric field is, E_y = kq_1/r^2 sin 45 degree = (9 times 10^9) (1.90 times 10^-6 C)/(0.792 m)^2 sin 45 degree = 19.28 times 10^3 j^hat The magnitude of the net electric field is, E = squareroot E^2_x + E^2_y = squareroot (58.51 times 10^3)^2 + (19.28 times 10^3)^2 = 61.6 times 10^3 N/C