Electrostatic Potential Energy of Two Point Charges and of a Dipole in an Electrostatic Field

Definition

• When two like charges are infinitely far apart their potential energy is zero, because no work has to be done to move one charge at an infinite distance from the other. Bringing them closer requires work against the force of repulsion, and since the electrostatic force is conservative this work is stored as potential energy.
• Electrostatic potential energy of a system of point charges is the total amount of work done to assemble the system of charges by bringing them from infinity to their present locations.

(a) Potential energy of a system of two point charges

Derivation

• Take charges $q_1$ and $q_2$ with position vectors $\vec{r_1}$ and $\vec{r_2}$, both initially at infinity, and calculate the work done by an external agency in bringing them to their places.
• Bringing $q_1$ to A: no work is done, because there is no external field to work against ($q_2$ is still at infinity). So $W_1 = 0$. This charge sets up a potential in space,

$$V_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r_1}$$

• Bringing $q_2$ from infinity to B, at a distance $r_{12}$ from $q_1$:

$$W_2 = (\text{potential at B due to } q_1)\times q_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$$

• This total work is stored as the potential energy of the configuration:

$$\boxed{\,U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}\,}$$

• Keep the signs: $U$ is positive for like charges and negative for unlike charges.

(b) Potential energy of a system of N point charges

• Bringing a third charge $q_3$ to C requires work against the fields of both $q_1$ and $q_2$:

$$W_3 = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{r_{13}}+\dfrac{q_2}{r_{23}}\right)q_3 = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1q_3}{r_{13}}+\dfrac{q_2q_3}{r_{23}}\right)$$

• Similarly, bringing $q_4$ to D,

$$W_4 = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1q_4}{r_{14}}+\dfrac{q_2q_4}{r_{24}}+\dfrac{q_3q_4}{r_{34}}\right)$$

• Continuing in this way, the electrostatic potential energy of a system of $N$ point charges is

$$\boxed{\,U = \dfrac{1}{4\pi\varepsilon_0}\sum_{\text{all pairs}}\dfrac{q_jq_k}{r_{jk}}\,}$$

• Count each pair once only: three charges give three terms, four charges give six.
• Worked standard result: for $-q$, $+Q$, $-q$ placed at equal spacings $d$ on a straight line, $U = \dfrac{1}{4\pi\varepsilon_0}\left(-\dfrac{2qQ}{d}+\dfrac{q^2}{2d}\right)$, and setting $U=0$ gives $Q:q = 1:4$.

(c) Potential energy of a single charge in an external electric field

• Here the field $\vec{E}$ is not produced by the charge whose energy we want; the external sources may be unknown, but the external field $E$ or the external potential $V$ is known. The field is assumed unaffected by the charge $q$, which is taken to be very small.
• If $V(\vec{r})$ is the external potential at a point P with position vector $\vec{r}$, then by definition the work done in bringing unit positive charge from infinity to P equals $V$. Hence the work in bringing a charge $q$ is $qV(\vec{r})$, and this is stored as its potential energy:

$$\boxed{\,\text{PE} = qV(\vec{r})\,}$$

(d) Potential energy of two charges in an external electric field

• Work in bringing $q_1$ from infinity to $\vec{r_1}$: $W = q_1V(\vec{r_1})$.
• Bringing $q_2$ to $\vec{r_2}$ requires work against the external field and against the field of $q_1$. Work against the external field is $q_2V(\vec{r_2})$; work against the field of $q_1$ is $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$, where $r_{12}$ is the distance between them. By the superposition principle these add.
• Total potential energy of the system:

$$\boxed{\,U = q_1V(\vec{r_1}) + q_2V(\vec{r_2}) + \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}\,}$$

• The third term is the one candidates lose marks by omitting.

(e) Potential energy of a dipole in an external field

Set-up and diagram

• A dipole of charges $-q$ and $+q$ separated by $2l$ is placed in a uniform electric field $\vec{E}$. The two equal and opposite forces $qE$ form a couple, so the net force is zero but the dipole experiences a torque that tends to rotate it. Draw the field lines horizontal, the dipole inclined at $\theta$, and the two forces $+qE$ and $-qE$ at the ends.

$$\vec{\tau} = \vec{p}\times\vec{E}\qquad\text{or}\qquad \tau = pE\sin\theta$$

Derivation of the energy

• To neutralise this torque an external torque $\tau_{\text{ext}}$ is applied, rotating the dipole in the plane of the paper from $\theta_0$ to $\theta$ without angular acceleration and at an infinitesimal angular speed. The work done by the external torque is

$$W = \int_{\theta_0}^{\theta}\tau_{\text{ext}}\,d\theta = \int_{\theta_0}^{\theta}pE\sin\theta\,d\theta = pE\left[-\cos\theta\right]_{\theta_0}^{\theta}$$

$$W = pE\left(\cos\theta_0 - \cos\theta\right)$$

• This work is stored as the potential energy of the system in the position where the dipole makes angle $\theta$ with the field:

$$U(\theta) - U(\theta_0) = pE\left(\cos\theta_0-\cos\theta\right)$$

The two standard choices of zero

• If the dipole is initially perpendicular to the field, $\theta_0 = \dfrac{\pi}{2}$, then $U(\theta) = pE\left(\cos 90^\circ - \cos\theta\right)$, i.e.

$$\boxed{\,U(\theta) = -pE\cos\theta\,}$$

• If the dipole is initially parallel to the field, $\theta_0 = 0$, then

$$\boxed{\,U(\theta) = pE\left(1-\cos\theta\right)\,}$$

• The zero of potential energy may be chosen as convenient, so always state which convention is being used.

Working with the torque and work formulae

• Maximum torque occurs at $\theta = 90^\circ$: $\tau_{max} = pE\sin 90^\circ = pE$. Torque is zero at $\theta = 0^\circ$ (stable) and $\theta = 180^\circ$ (unstable).
• Work done by an external agent in turning the dipole from $\theta_1$ to $\theta_2$:

$$W = pE\left(\cos\theta_1-\cos\theta_2\right)$$

• For turning through $180^\circ$ from $\theta = 0^\circ$, $W = pE(1+1) = 2pE$. For $N$ molecules all turned together, multiply by $N$.