A charge of +3 C is at the origin. When charge Q is placed at 2 m along the positive x‑axis, the electric field at 2 m along the negative ´‑axis becomes zero. What is the value of Q?
a. ‑3 C
b. ‑6 C
c. ‑9 C
d. ‑12 C
please justify your answer thanks!
I will be happy to critique your thinking.
To find the value of Q, we can use the principle that the electric field due to a point charge is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge.
Given that a charge of +3 C is at the origin, we can determine the electric field at any point by applying the formula for electric field strength:
Electric field (E) = k * |Q| / r^2
In this case, we know that at a distance of 2 m along the negative y-axis from the origin, the electric field becomes zero. This means that the electric field due to charge Q is equal in magnitude but opposite in direction to the electric field due to the +3 C charge at the origin.
Since the electric field due to the +3 C charge at the origin is positive, the electric field due to charge Q must be negative to cancel it out and result in a net electric field of zero.
Therefore, to find the value of Q, we equate the magnitudes of the electric fields:
k * |3 C| / (2 m)^2 = k * |Q| / (2 m)^2
Simplifying this equation, we can cancel out the constant terms and solve for Q:
3 C = -Q
Q = -3 C
Therefore, the value of Q is -3 C (option a).
We arrived at this answer by using the principle of electric field cancellation and setting the magnitudes of the two electric fields equal to each other.