Problem 10-38 Addition of Electric fields - 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

Correct.

The magnitude of the field $E_1$ due to $q_1$ at the $\text{point A}$ is:

(A)   $|E_1| = k|q_1|/r_1$

(B)   $|E_1| = k|q_1|/r_1{^2}$

(C)    $|E_1| = k|q_1|r_1{^2}$