Example 5.4: Electric potential due to point charges

Question: Suppose that three point charges, $q_a$, $q_b$, and $q_c$, are arranged at the vertices of a right-angled triangle, as shown in the diagram. What is the absolute electric potential of the third charge if $q_a=-6.0\,\mu{\rm C}$, $q_b=+4.0\,\mu{\rm C}$, $q_c = +2.0\,\mu{\rm C}$, $a=4.0$m, and $b=3.0$m? Suppose that the third charge, which is initially at rest, is repelled to infinity by the combined electric field of the other two charges, which are held fixed. What is the final kinetic energy of the third charge?
 
Solution: The absolute electric potential of the third charge due to the presence of the first charge is

$$V_a = k_e\,\frac{q_a}{c} = (8.988\times 10^9)\,\frac{(-6\times 10^{-6})}{(\sqrt{4^2+3^2})}= - 1.08 \times 10^4\,{\rm V},$$

where use has been made of the Pythagorean theorem. Likewise, the absolute electric potential of the third charge due to the presence of the second charge is

$$V_b = k_e\,\frac{q_b}{b} = (8.988\times 10^9)\,\frac{(4\times 10^{-6})}{(3)}=1.20\times 10^4\,{\rm V}.$$

The net absolute potential of the third charge $V_c$ is simply the algebraic sum of the potentials due to the other two charges taken in isolation. Thus,

$$V_c = V_a + V_b =1.20\,\times 10^3\,{\rm V}.$$

The change in electric potential energy of the third charge as it moves from its initial position to infinity is the product of the third charge, $q_c$, and the difference in electric potential ($-V_c$) between infinity and the initial position. It follows that

$${\mit\Delta}P = -q_c\,V_c = -(2\times 10^{-6})\,(1.2\times 10^3)=-2.40\times 10^{-3}\,{\rm J}.$$

This decrease in the potential energy of the charge is offset by a corresponding increase ${\mit\Delta}K= -{\mit\Delta}P$ in its kinetic energy. Since the initial kinetic energy of the third charge is zero (because it is initially at rest), the final kinetic energy is simply

$$K = {\mit\Delta}K = -{\mit\Delta}P = 2.40\times 10^{-3}\,{\rm J}.$$