Question: What is the electric field (E) at a distance of 2 meters from a point charge of 5 microcoulombs?

Options:

1. 1125 N/C
2. 450 N/C
3. 225 N/C
4. 900 N/C

Correct Answer: 1125 N/C

Solution:

Using Coulomb\'s law, E = k * |q| / r^2, where k = 8.99 x 10^9 N m^2/C^2, q = 5 x 10^-6 C, and r = 2 m. E = (8.99 x 10^9) * (5 x 10^-6) / (2^2) = 1125 N/C.