Problem 10-38 Addition of Electric fields - Part 3

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.

\( |E_1| = k|q_1|/r_1{^2}\)

\(|E_1| = k|q_1|/r_1{^2}  = (8.99 \times10^9  )(3.6 \times 10^{-8} )/(3.8 \times 10^{-6} )^2 = 2.24 \times 10^{13}\; N/C\)

Do you see where the number \(3.8 \times 10^{-6}\) came from?    No

The direction of the field \(E_2\) due to \(q_2\) at the \(\text{point A}\) is:

Two possible electric field directions; A, B.