Problem 10-38 Addition of Electric fields - Part 4

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.

\(|E_2| = k|q_2|/r_2{^2} = (8.99 \times 10^9)(5.2 \times 10^{-8} )/(2.5 \times 10^{-6})2 = 7.48 \times 10^{13}\; N/C\)

The net field at \(A\) is:

(A)   \(7.48 \times 10^{13} + 2.24 \times 10^{13} = 9.72 \times 10^{13}\; N/C\; \text{to the right}\)

(B)   \(7.48 \times 10^{13} - 2.24 \times 10^{13} = 5.24 \times 10^{13}\; N/C\; \text{ to the left}\)

(C)   \(7.48 \times 10^{13} - 2.24 \times 10^{13} = 5.24 \times 10^{13}\; N/C\; \text{ to the right}\)

(D)   \(7.48 \times 10^{13} + 2.24 \times 10^{13} = 9.72 \times 10^{13}\; N/C\; \text{ to the left}\)