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

We can find the electric field due to the +15-nC charge and the electric field due to the −20-nC charge separately, and then find the net electric field at the origin by adding these vectors.

Using the equation for electric field due to a point charge, we have:

E1 = k*q1/r1^2 = (9.0×10^9 N*m^2/C^2)*(+15×10^(-9) C)/(1.5 m)^2 = 150 N/C

(Note that k is Coulomb's constant.)

The electric field due to the −20-nC charge is directed along the negative y-axis and has a magnitude of:

E2 = k*q2/r2^2 = (9.0×10^9 N*m^2/C^2)*(-20×10^(-9) C)/(2.0 m)^2 = 45 N/C

(Note that we use the magnitude of the charge, since the electric field is a vector and we are only interested in its magnitude at this point.)

To add the electric fields, we can use vector addition. The x-component of E1 is zero (since it is directed along the x-axis), and the y-component is 150 N/C. The y-component of E2 is −45 N/C (since it is directed opposite to the positive y-axis), and the x-component is zero. Therefore, the net electric field is:

E_net = sqrt[(E1,y + E2,y)^2 + 0^2] = sqrt[(150 N/C − 45 N/C)^2 + 0^2] = 75 N/C

Thus, the magnitude of the electric field at the origin is 75 N/C, which corresponds to choice C.

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