Problem 10-38 Addition of Electric fields - Part 4 - C
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
\( |E_1| = k|q_1|/r_1{^2} \\ |E_1| = k|q_1|/r_1{^2} = 2.24 \times 10^{13} \; N/C\)
\( |E_2| = k|q_2|/r_2{^2} = 7.48 \times 10^{13}\; N/C\)
Correct.
Net \(E\) at \(A = 5.2 \times10^{13}\; N/C\) to the right (answer to part (a))
We now examine the same charge configuration but evaluate the field at \(\text{point B}\)
Which figure below has the correct direction for the fields \(E_1\) and \(E_2\) at \(B\)?
