Question

particle (charge = +6 nC) is located at the origin (* = 0), and a second particle (charge =+3 nC) is placed on the x axis at x =2 m. What is the electric potential (in V) at a point of x = 3 m? a. 45 b. 20 d. 50 e. 30

Answer 1

Potential created by the first charge at x=3 m:

$$V_1=\frac{kq_1}{x-r_1}.$$

Potential created by the second charge at x=3m:

$$V_2=\frac{kq_2}{x-r_2}.$$

The total potential:

$$V=V_1+V_2,\\\space\\ V=k\bigg(\frac{q_1}{x-r_1}+\frac{q_2}{x-r_2}\bigg),\\\space\\ V=9\cdot10^9\bigg(\frac{6\cdot10^{-9}}{3-0}+\frac{3\cdot10^{-9}}{3-2}\bigg)=45\text{ V}.$$