A +15-nC point charge is placed on the x axis at x = 1.5 m,and a -20-nC charge is placed on the y axis at y = -2.0m.What is the magnitude of the electric field at the origin?
A)105 N/C
B)15 N/C
C)75 N/C
D)45 N/C
E)60 N/C

Question

Quizplus · Accepted Answer

To find the electric field at the origin, we first need to find the electric field at the origin due to each point charge, and then use vector addition to find the total electric field.

The electric field due to the +15 nC charge at the origin is:
E1 = kq1/r1^2 = (9x10^9 Nm^2/C^2)(15x10^-9 C)/(1.5 m)^2 = 60 N/C (directed along the negative x-axis)

The electric field due to the -20 nC charge at the origin is:
E2 = kq2/r2^2 = (9x10^9 Nm^2/C^2)(20x10^-9 C)/(2.0 m)^2 = 22.5 N/C (directed along the positive y-axis)

Now we need to find the net electric field at the origin:
E = sqrt(E1^2 + E2^2) = sqrt((60 N/C)^2 + (22.5 N/C)^2) = 75 N/C

Therefore, the magnitude of the electric field at the origin is 75 N/C, which is answer choice C.

A +15-NC Point Charge Is Placed on the X Axis