# 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:

(A)

(B)