A point charge of 3.0 µC is located on the y-axis at y = 4.0 m, and another point charge of 6.0 µC is located on the x-axis at x = 5.0 m. What is the magnitude of the electric field at the position (5.0 m, 4.0 m)? A) 2.7 × 10³ N/C B) 3.5 × 10³ N/C C) 2.7 × 10^-3 N/C D) 3.5 × 10^-3 N/C E) 4.5 × 10³ N/C

Question

Quizplus · Accepted Answer

We can use the formula for electric field due to a point charge:
E = kq/r^2
where k is Coulomb's constant, q is the charge, and r is the distance from the charge. To find the electric field at the given position, we need to find the electric field due to each charge and then add them up vectorially.

For the charge of 3.0 µC at y = 4.0 m:
r = sqrt((5.0 - 0)^2 + (4.0 - 4.0)^2) = 5.0 m
E1 = (9.0 × 10^9 N m^2/C^2) * (3.0 × 10^-6 C) / (5.0 m)^2 = 54 N/C (pointing along the negative y-axis)

For the charge of 6.0 µC at x = 5.0 m:
r = sqrt((5.0 - 5.0)^2 + (4.0 - 0)^2) = 4.0 m
E2 = (9.0 × 10^9 N m^2/C^2) * (6.0 × 10^-6 C) / (4.0 m)^2 = 33.75 N/C (pointing along the positive x-axis)

The net electric field is the vector sum of these two:
E_net = sqrt(E1^2 + E2^2) = sqrt((54 N/C)^2 + (33.75 N/C)^2) = 63 N/C

Therefore, the magnitude of the electric field at the given position is 63 N/C, which is closest to choice B (3.5 × 10^13 N/C).

A Point Charge of 3