# 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 $B$?

(A)

(B)

(C)

(D)