The electric field produced by a point charge Q = 5 μC at a distance r = 0.20 m is E = kQ/r^2. What is E (in N/C)? (k ≈ 8.99×10^9)

The electric field produced by a point charge Q = 5 μC at a distance r = 0.20 m is E = kQ/r^2. What is E (in N/C)? (k ≈ 8.99×10^9)

• A. About 1.12 × 10^6 N/C
• B. About 2.00 × 10^6 N/C
• C. About 5.00 × 10^5 N/C
• D. About 1.00 × 10^4 N/C

Answer: (A)

Electric field from a point charge scales as E = k Q / r^2. Convert the given values: Q = 5 μC = 5×10^-6 C and r = 0.20 m, so r^2 = 0.04 m^2. Then E = (8.99×10^9) × (5×10^-6) / 0.04 = 4.495×10^4 / 0.04 = 1.12375×10^6 N/C. So about 1.12×10^6 N/C. The large field is due to both the sizable charge (in microcoulombs) and the short distance. If the distance were larger or the charge smaller, the field would be much weaker.

Electric field from a point charge scales as E = k Q / r^2. Convert the given values: Q = 5 μC = 5×10^-6 C and r = 0.20 m, so r^2 = 0.04 m^2. Then E = (8.99×10^9) × (5×10^-6) / 0.04 = 4.495×10^4 / 0.04 = 1.12375×10^6 N/C. So about 1.12×10^6 N/C. The large field is due to both the sizable charge (in microcoulombs) and the short distance. If the distance were larger or the charge smaller, the field would be much weaker.