Problem 10-38 Addition of Electric fields - Part 6

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.

Diagram indicating magnitude and direction of charge.

\(|E_1| = k|q_1|/r_3{^2}  = (8.99 \times 10^9  )(3.6 \times 10^{-8} )/(8.2 \times 10^{-6} )^2 = 4.81 \times 10^{12}\; N/C\; \text{ to the left}\)

Do you see where the \(8.2 \times 10^{-6}\) came from?   No?
\(|E_2| = k|q_1|/r_4{^2}  = (8.99 \times 10^9  )(5.2 \times 10^{-8} )/(1.9 \times 10^{-6} )^2 = 1.29\times 10^{14}\; N/C\; \text{ to the left}\)

Before continuing calculate the net field at \(B.\)

Continue.