A point charge q = 5 μC is located at r = 0.2 m away from a point of interest. Assuming vacuum permittivity, what is the magnitude of the electric field due to this charge at that point?

A point charge q = 5 μC is located at r = 0.2 m away from a point of interest. Assuming vacuum permittivity, what is the magnitude of the electric field due to this charge at that point?

• A. Approximately 1.12×10^9 N/C
• B. 1.12×10^3 N/C
• C. 1.12×10^12 N/C
• D. Approximately 1.12×10^6 N/C

Answer: (D)

The electric field from a point charge in vacuum falls off with the square of the distance: E = k q / r^2, where k = 1/(4π ε0) ≈ 8.99×10^9 N·m^2/C^2. With a charge of 5 μC (5×10^-6 C) at a distance of 0.2 m, compute E = (8.99×10^9)(5×10^-6)/(0.2)^2. The numerator is about 4.495×10^4, and dividing by 0.04 gives roughly 1.12×10^6 N/C. The magnitude is about 1.12×10^6 N/C, which reflects how a larger charge increases the field and a larger distance decreases it.

The electric field from a point charge in vacuum falls off with the square of the distance: E = k q / r^2, where k = 1/(4π ε0) ≈ 8.99×10^9 N·m^2/C^2. With a charge of 5 μC (5×10^-6 C) at a distance of 0.2 m, compute E = (8.99×10^9)(5×10^-6)/(0.2)^2. The numerator is about 4.495×10^4, and dividing by 0.04 gives roughly 1.12×10^6 N/C. The magnitude is about 1.12×10^6 N/C, which reflects how a larger charge increases the field and a larger distance decreases it.