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 $\vec { E }$ field at the 30 cm mark? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

The electric field ($\vec{E}$) due to a point charge is given by $E = k \frac{|q|}{r^2}$, where $k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, $q$ is the charge, and $r$ is the distance from the charge to the point of interest. At the 30 cm mark, the distance from the +5.0-μC charge is 30 cm (0.30 m) and from the -4.0-μC charge is 20 cm (0.20 m). The electric fields due to each charge at the 30 cm mark can be calculated and then added vectorially, considering their directions (towards the negative charge and away from the positive charge). The net electric field is found to be $1.4 \times 10^{16} \, \text{N/C}$, after calculating and summing the contributions from both charges correctly.

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