Problem 10-38 Addition of Electric fields - Part 5 - D

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.

\( |E_1| = k|q_1|/r_1{^2} \\ |E_1| = k|q_1|/r_1{^2} = 2.24 \times 10^{13} \; N/C\)

Diagram indicating electric field moving to the right.

\( |E_2| = k|q_2|/r_2{^2} = 7.48 \times 10^{13}\; N/C\)

Diagram of charge configuration.

No. Both of the directions are incorrect. No. To determine the direction of the field at \(B\) due to \(q_1\) we imagine placing a small test charge at \(B\) and observing the direction of the force on it. The charge should be Positive or Negative? Choose one.