Question

A –8.0-µC charge is located 0.45 m to the left of a +6.0-µC charge. What is the magnitude and direction of the electrostatic force on the positive charge?

Answer 1

Answer:

2.133 N to the left

Explanation:

We can represent the situation with the following equation:

Then, the magnitude can be calculated as:

$\begin{gathered} F=k\frac{q_1q_2}{r^2} \\ \\ \text{ Where k = 9 x 10}^9\text{ N m}^2\text{ /C}^2 \end{gathered}$

Where q1 and q2 are the charges and r is the distance. So, replacing q1 = 8 x 10^(-6) C, q2 = 6.0 x 10^(-6) C, and r = 0.45 m, we get:

$F=(9\times10^9)\frac{(8\times10^{-6})(6\times10^{-6})}{0.45^2}=2.133\text{ N}$

Therefore, the magnitude of the electrostatic force is 2.133 N and the direction is to the left.