Problem 10-38 Addition of Electric fields - Part 3
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
\( |E_1| = k|q_1|/r_1{^2}\)
\(|E_1| = k|q_1|/r_1{^2} = (8.99 \times10^9 )(3.6 \times 10^{-8} )/(3.8 \times 10^{-6} )^2 = 2.24 \times 10^{13}\; N/C\)
Do you see where the number \(3.8 \times 10^{-6}\) came from? No
The direction of the field \(E_2\) due to \(q_2\) at the \(\text{point A}\) is:
