Problem 10-38 Addition of Electric fields - Part 2 - B
Two negative charges have locations as shown in the Figure. Charge \(q_1\) is \(-3.6\times 10^{-8} C;\) charge \(q_2\) is \(-5.2 \times10^{-8} C.\) What is the electric field (magnitude and direction) at (a) \(\text{point A}\)? (b) \(\text{point B}\)?
Accumulated Solution
\( |E_1| = k|q_1|/r_1{^2}\)
Correct.
Before continuing, calculate the value of \(|E_1| .\) The value of \(k\) is \(8.99 \times 10^9\; Nm^2/C^2\)