Problem 10-38 Addition of Electric fields - B

Diagram of two negative charges.

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

Diagram B of direction of field.

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)    \(|E_1| = k|q_1|r_1{^2}\)