Physics · Ch 8 — Electrostatics

Potential Energy of a System of Two Point Charges

8.6.1

Potential Energy of a System of Two Point Charges

Bring the first charge $q_1$ in from infinity to its final position $\vec{r}_1$ . Since the second charge $q_2$ is still out at infinity at this stage, there is no field yet to do work against, so this first step costs zero work: $W_1=0$ . Once in place, $q_1$ produces its own potential everywhere in space, in particular at $q_2$ 's eventual location: $V_1=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1}{r_1}$ .

Now bring the second charge $q_2$ in from infinity to its final position $\vec{r}_2$ , a distance $r_{12}$ from $q_1$ . This step requires work against $q_1$ 's own field, equal to (the potential due to $q_1$ at that location) $\times\,q_2$ : $W_2=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r_{12}}$ . …

Figure 8.16Fig. 8.16: System of two point charges

Fig. 8.16 — Fig. 8.16: System of two point charges

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Two point charges $q_1$ and $q_2$ , located at position vectors $\vec{r}_1$ and $\vec{r}_2$ measured from a common origin O, joined by a straight line of length $r_{12}$ (their mutual separation) -- the basic vector geometry used to write the two-charge assembly work as $W_2=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}}$ , which becomes the potential energy of the pair once bot …

Misc Ex.10Example 8.10: PE of a 5 nC and -2 nC charge pair, 18 cm apart

Worked out. Charges $q_1=5$ nC $=5\times10^{-9}$ C at $(2\text{ cm},0,0)$ and $q_2=-2$ nC $=-2\times10^{-9}$ C at $(20\text{ cm},0,0)$ , so $r=18\text{ cm}=18\times10^{-2}$ m apart. $U=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r}=9\times10^9\times\dfrac{5\times10^{-9}\times(-2\times10^{-9})}{18\times10^{-2}}=-5\times10^{-7}\,\text{J}=-0.5\,\mu\text{J}$ -- negative, as expected for a pair of unlike charges, meaning their own mutual attraction did the work of bringing them to thi …

Misc Ex.12Example 8.12: Increase in PE bringing two equal charges to 30 cm apart

Worked out. Two equal charges $q_1=q_2=3\times10^{-5}$ C are brought from infinity (zero PE) to a separation of $r=0.30$ m. The increase in PE equals the present PE: $U=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r}=9\times10^9\times\dfrac{(3\times10^{-5})^2}{0.3}=27$ J -- a large, positive value since both charges are like-signed, illustrating how much work is needed to force two sizeable like charges into close proximity. …

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