Final answer:
The second charge Q must be positive to ensure that the resultant electric field at the origin is 45 N/C in the −x-direction. By determining the contributions of both charges and setting them equal to the desired electric field, we can solve for the magnitude of charge Q.
Step-by-step explanation:
To determine the sign and magnitude of the point charge Q that would result in an electric field of 45 N/C in the −x-direction at the origin, consider each point charge's contribution to the electric field at that point. The electric field due to a point charge is given by E = k |q| / r2, where E is the electric field, k is Coulomb's constant (8.99 × 109 N m2/C2), q is the charge, and r is the distance from the charge to the point where the electric field is measured.
The −3.00-nC charge at x = 1.20 m contributes to the electric field at the origin in the +x-direction since it is a negative charge. The second charge Q at x = −0.600 m will contribute to the electric field in the −x-direction if it is positive. We need Q's contribution to be strong enough to not only cancel out the contribution of the −3.00-nC charge but also to provide the net electric field of 45 N/C in the −x-direction.
As the desired resultant electric field is in the −x-direction, it is clear that the charge Q must be positive. To find the magnitude of Q, we need to set up an equation that sums the individual electric fields from both charges at the origin and solves for Q, ensuring that the net electric field is −45 N/C:
k (−3.00 × 10−6 C) / (1.20 m)2 + k Q / (0.600 m)2 = −45 N/C. Solving for Q gives us the magnitude of the positive charge required.