Example 5.3: Electric potential due to point charges

Question:  A particle of charge $q_1=+6.0\,\mu{\rm C}$ is located on the $x$-axis at the point $x_1=5.1\,{\rm cm}$. A second particle of charge $q_2=-5.0\,\mu{\rm C}$ is placed on the $x$-axis at $x_2=-3.4\,{\rm cm}$. What is the absolute electric potential at the origin ($x=0$)? How much work must we perform in order to slowly move a charge of $q_3=-7.0\,\mu{\rm C}$ from infinity to the origin, whilst keeping the other two charges fixed?
 
Solution: The absolute electric potential at the origin due to the first charge is

$$V_1 = k_e\,\frac{q_1}{x_1} = (8.988\times 10^9)\,\frac{(6\times 10^{-6})}{(5.1\times 10^{-2})}= 1.06 \times 10^6\,{\rm V}.$$

Likewise, the absolute electric potential at the origin due to the second charge is

$$V_2 = k_e\,\frac{q_2}{\vert x_2\vert} = (8.988\times 10^9)\,... ...imes 10^{-6})}{(3.4\times 10^{-2})}=-1.32\times 10^6\,{\rm V}.$$

The net potential $V$ at the origin is simply the algebraic sum of the potentials due to each charge taken in isolation. Thus,

$$V = V_1 + V_2 = -2.64\times 10^5\,{\rm V}.$$

The work $W$ which we must perform in order to slowly moving a charge $q_3$ from infinity to the origin is simply the product of the charge and the potential difference $V$ between the end and beginning points. Thus,

$$W = q_3\, V = (-7\times 10^{-6})\,(-2.64\times 10^5) = 1.85\,{\rm J}.$$