Problem 10-38 Addition of Electric fields - Part 2 - C
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
No. The equation for the electric field is just another form of Coulamb's Law for the force between 2 charges \((F = kqq'/r^2)\) except the field is defined as the "force per unit charge" (i.e.divide through by one of the charges. Look at the choices again.