Problem 10-38 Addition of Electric fields - Part 4 - A
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
\( |E_1| = k|q_1|/r_1{^2} \\ |E_1| = k|q_1|/r_1{^2} = 2.24 \times 10^{13} \; N/C\)
\( |E_2| = k|q_2|/r_2{^2} = 7.48 \times 10^{13}\; N/C\)
No. The two field vectors point in opposite directions.