Potential Energy in an External Field

What is different here

In the previous section the charges themselves made the field. Here the field $\vec E$ (or the potential $V$) is produced by external sources that are not part of the system, and are assumed to be undisturbed by the charge we place in it. We want the energy of the charge in that field, not the energy of the sources.

A single charge

$$U=qV(\vec r)$$

• $V(\vec r)$ is the external potential at the point, and the work done in bringing $q$ from infinity to $\vec r$ is $qV$, stored as potential energy.
• $V$ here must exclude the potential of $q$ itself, which is infinite at its own location.
• The electron volt follows: an electron taken through $\Delta V=1\ \text{V}$ gains $1.6\times10^{-19}\ \text{J}=1\ \text{eV}$.
• A charge moved round a closed loop in this field gains nothing: the work done over the closed rectangle $abcda$ between capacitor plates is zero, because the electrostatic field is conservative.

Two charges in an external field

Bring $q_1$ to $\vec r_1$: work $=q_1V(\vec r_1)$. Bring $q_2$ to $\vec r_2$: work against the external field is $q_2V(\vec r_2)$, and work against the field of $q_1$ is $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$. Adding,

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

• Written with the potentials at the two locations called $V_A$ and $V_B$, the same expression reads $W=q_1V_A+q_2V_B+\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}$.
• The mutual term is unchanged by the presence of the external field; the external field only adds the first two terms.
• Textbook illustration: $7\ \mu\text{C}$ and $-2\ \mu\text{C}$ at $(\pm9\ \text{cm},0,0)$ have mutual energy $-0.7\ \text{J}$; in an external field $E=A/r^{2}$ with $A=9\times10^{5}\ \text{N C}^{-1}\text{m}^{2}$, the total electrostatic energy becomes $70-20-0.7=49.3\ \text{J}$.

Dipole in a uniform external field — the 5-mark derivation

In a uniform field the two charges of a dipole feel equal and opposite forces $qE$, so the net force is zero, but the forces are not collinear and produce a torque:

$$\tau=qE(2a\sin\theta)=pE\sin\theta,\qquad \vec\tau=\vec p\times\vec E$$

To rotate the dipole quasi-statically an external torque just balances this. Work done in turning through $d\theta$:

$$dW=\tau\,d\theta=pE\sin\theta\,d\theta$$

$$W=\int_{\theta_1}^{\theta_2}pE\sin\theta\,d\theta=pE\left[\cos\theta_1-\cos\theta_2\right]$$

This work is stored as potential energy. Choosing $U=0$ at $\theta_1=90^\circ$ — the natural choice, because there the works done against the field in bringing $+q$ and $-q$ into place are equal and opposite and cancel — gives

$$U(\theta)=pE\left[\cos90^\circ-\cos\theta\right]=-pE\cos\theta=-\vec p\cdot\vec E$$

Equilibrium and extreme values

• Minimum potential energy is $U_{\min}=-pE$, at $\theta=0^\circ$, with $\vec p$ parallel to $\vec E$.
• Work to rotate a dipole initially parallel to the field through $90^\circ$: $W=pE(\cos0^\circ-\cos90^\circ)=pE$. From $90^\circ$ to $180^\circ$ the work is again $pE$; the full flip from $0^\circ$ to $180^\circ$ costs $2pE$.
• Numerical pattern: $p=6\times10^{-7}\ \text{C m}$ in $E=10^{4}\ \text{N C}^{-1}$ with $\vec p$ parallel to $\vec E$ gives $U=-pE=-6\times10^{-3}\ \text{J}$.
• Related Chapter-1 result reused in these questions: an ideal dipole is two equal and opposite charges separated by a vanishingly small distance with $p=q(2a)$ finite; a water molecule is the standard example. Net force in a uniform field is zero.

Charge brought up to an infinite charged sheet

For an infinite plane sheet of uniform density $+\sigma$, the field is $\dfrac{\sigma}{2\varepsilon_0}$ everywhere, independent of distance. Then

$$V_P-V_\infty=-\int_\infty^{r}\dfrac{\sigma}{2\varepsilon_0}\,dr=-\dfrac{\sigma}{2\varepsilon_0}\left(r-\infty\right)\to\infty$$

so $W=q(V_P-V_\infty)$ is infinite: an infinite amount of work is required, because the field of an infinite sheet does not die away at large distance. The point of the question is exactly that the usual "potential zero at infinity" convention fails for an infinite charge distribution.

Diagram to describe in words

A dipole drawn as $+q$ and $-q$ joined by a rod of length $2a$, tilted at angle $\theta$ to a set of horizontal, equally spaced field lines representing uniform $\vec E$. Show the force $qE$ on $+q$ along the field and $qE$ on $-q$ against it, forming a couple, with the perpendicular distance between their lines of action marked $2a\sin\theta$.

Standard results