A +5.0-μC point charge is placed at the 0 cm mark of a meter stick and a -4.0-μC charge is placed at the 50 cm mark.What is the net electric field at the 30 cm mark? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²) A)4.0 × 10⁵ N/C B)5.0 × 10⁵ N/C C)9.0 × 10⁵ N/C D)1.4 × 10⁶ N/C

Question

Quizplus · Accepted Answer

- Calculate the electric field due to the +5.0-μC charge at the 30 cm mark:
E1 = kq1 / r1^2 = (8.99 × 10^9 N∙m^2/C^2)(5.0 × 10^-6 C) / (0.3 m)^2 = 1.11 × 10^6 N/C (towards the left)
- Calculate the electric field due to the -4.0-μC charge at the 30 cm mark:
E2 = kq2 / r2^2 = (8.99 × 10^9 N∙m^2/C^2)(-4.0 × 10^-6 C) / (0.2 m)^2 = -4.49 × 10^6 N/C (towards the right)
- Calculate the net electric field at the 30 cm mark by finding the vector sum of E1 and E2 (the fields are in opposite directions):
Enet = |E1 + E2| = |1.11 × 10^6 N/C - 4.49 × 10^6 N/C| = 3.38 × 10^6 N/C
The answer is the closest to D: 1.4 × 10^16 N/C (scientific notation format).

A +50-μC Point Charge Is Placed at the 0 Cm Mark