Potential energy in an external field

This sits inside the CBSE Class 12 Physics chapter "Electrostatic Potential and Capacitance" (Unit II, Electrostatics), directly after the potential energy of a system of point charges. The prescribed scope is exactly two formulas — $U=qV$ for a single charge in an external field, and $U=-\vec p\cdot\vec E$ for a dipole in a uniform external field — together with reading off the stable/unstable orientations and connecting to the torque $\tau=pE\sin\theta$ from the electric-dipole topic that came before it. Nothing beyond this (no energy of a dipole in a non-uniform field, no higher multipoles) is examined at this level, so this article stays deliberately narrow and goes deep on the one result that carries almost all the marks.

Start with a single point charge $q$ placed at some location inside a field produced by other, external charges — the field and its potential $V$ at that point already exist before $q$ arrives. By the very definition of electric potential as "potential energy per unit charge," the potential energy this charge acquires by being there is

where: $q$ is the test/placed charge (coulombs, can be positive or negative), $V$ is the electric potential at that point due to the external sources only (volts), and $U$ comes out in joules. A positive charge parked at a high-potential point has large positive $U$ (it "wants" to move to lower potential, the way a ball wants to roll downhill); a negative charge at that same point has negative $U$ and is instead pulled toward higher potential. This single relation is really just $V=U/q$ rearranged — nothing new is being derived, only reinterpreted. Now replace the single charge with a rigid dipole: two equal and opposite charges $+q$ and $-q$ separated by a small fixed distance $2a$, forming a dipole moment $\vec p$ of magnitude $p=q(2a)$ pointing from $-q$ to $+q$. Place this dipole in a uniform external field $\vec E$, with its axis making angle $\theta$ with the field direction, and add up the potential energy of each end separately using $U=qV$ for each charge. The uniform field means the potential varies linearly with position along $\vec E$, and carrying through that bookkeeping (done in full below) collapses to a strikingly compact result:

where: $p$ is the dipole moment magnitude (C·m), $E$ is the uniform external field strength (N/C or V/m), and $\theta$ is the angle between $\vec p$ and $\vec E$. The physical reading is that $U$ depends on orientation alone — swing the dipole around at a fixed location in a uniform field and its energy oscillates between $-pE$ and $+pE$ as $\theta$ runs from $0°$ to $180°$, tracing out a cosine curve. There is no dependence on where the dipole sits, only on how it is tilted — a direct consequence of the field being uniform (translating the dipole sideways changes nothing, since both charges feel the same $E$).

Board level - dipole energy at a fixed angle

A dipole of moment $p=5\times10^{-9}\ \text{C}\,\cdot\,\text{m}$ sits in a uniform field of magnitude $E=4\times10^{4}\ \text{N/C}$, at $\theta=60°$ to the field. Find its potential energy.

Step 1. Write the formula: $U=-pE\cos\theta$.

Step 2. Substitute the given values: $p=5\times10^{-9}\ \text{C}\,\cdot\,\text{m}$, $E=4\times10^{4}\ \text{N/C}$, $\cos 60°=0.5$.

Step 3. Compute in stages: $pE = (5\times10^{-9})(4\times10^{4}) = 2\times10^{-4}\ \text{J}$, then $U=-(2\times10^{-4})(0.5)$.

Answer: $U = -1\times10^{-4}\ \text{J}$. The negative sign says the dipole is closer to its stable orientation ($\theta=0°$) than its unstable one ($\theta=180°$) — a $60°$ tilt is "on the aligned side" of perpendicular.

Competitive level - work done rotating a dipole between two orientations

A dipole of moment $p=2\times10^{-8}\ \text{C}\,\cdot\,\text{m}$ starts aligned with a uniform field of magnitude $E=10^{5}\ \text{N/C}$ ($\theta=0°$) and is rotated by an external agent to $\theta=90°$. Find the work done by the external agent, and check it against the work done by the field itself.

Step 1. The work done by an external agent equals the change in potential energy (this holds because the process is quasi-static and the dipole's kinetic energy stays zero): $W_{\text{ext}} = U_{90°} - U_{0°}$.

Step 2. Evaluate the two energies. At $\theta=0°$: $U_{0°} = -pE\cos0° = -pE = -(2\times10^{-8})(10^{5}) = -2\times10^{-3}\ \text{J}$. At $\theta=90°$: $U_{90°} = -pE\cos90° = 0$.

Step 3. Subtract: $W_{\text{ext}} = 0-(-2\times10^{-3}) = 2\times10^{-3}\ \text{J}$.

Step 4. Cross-check with energy conservation. The field does work $W_{\text{field}} = -\Delta U = -2\times10^{-3}\ \text{J}$ on the dipole — negative, because the field is being fought as the dipole is dragged away from its comfortable, aligned orientation, exactly consistent with $\theta=0°$ being the stable minimum.

Answer: the external agent does $+2\times10^{-3}\ \text{J}$ of work; the field does $-2\times10^{-3}\ \text{J}$ on the dipole, and the two are equal and opposite as energy conservation demands. Timing note: this is a two-line calculation once you trust $W=\Delta U$ — under a minute if you don't re-derive the formula each time. The one place students lose time is second-guessing the sign of $W_{\text{ext}}$; settle it by physical reasoning first (moving away from the stable orientation always costs positive external work), then let the algebra confirm it.

• Dropping the minus sign in $U=-pE\cos\theta$. Without it, $\theta=0°$ would read as the maximum energy and the whole stable/unstable labelling flips — a very easy way to lose marks on an otherwise correct calculation.
• Treating "zero torque" and "zero potential energy" as the same idea. Torque $\tau=pE\sin\theta$ vanishes at both $\theta=0°$ and $\theta=180°$ (both are equilibria), but $U$ is at its two different extremes there, not zero at either — $U=0$ happens at $\theta=90°$, where the torque is instead at its own maximum.
• Assuming $\theta=90°$ is some kind of equilibrium because $U=0$ there. It is not — a dipole released at $\theta=90°$ feels the full torque $\tau=pE$ and immediately starts rotating toward $\theta=0°$.
• Confusing the potential energy of a dipole in a field ($U=-\vec p\cdot\vec E$, a scalar describing the dipole's situation) with the potential due to a dipole at some field point ($V=\frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}$, a completely different quantity from the earlier topic) — the shared symbols $p$, $\theta$ invite mixing them up.
• Forgetting that $U=-\vec p\cdot\vec E$ applies to a uniform field only; the compact form relies on the linear potential drop used in Step 3 of the derivation, which does not hold in a non-uniform field (a topic outside this syllabus).

• State and correctly apply $U=qV$ for a single charge sitting in an external potential.
• Derive $U=-\vec p\cdot\vec E=-pE\cos\theta$ for a dipole in a uniform field, starting from $U=qV$ applied to each charge.
• Identify, without recomputation, the angles of stable equilibrium ($\theta=0°$), unstable equilibrium ($\theta=180°$), and zero potential energy ($\theta=90°$), and explain why only the first two are equilibria.
• Compute the work done rotating a dipole between two orientations as $W=\Delta U$, and check the sign against physical intuition (moving away from alignment costs positive work).
• Distinguish this dipole-in-a-field energy from both the potential due to a dipole and the mutual potential energy of a system of charges, despite the overlapping symbols.

Ask the tutor: want to see how this same $U=-pE\cos\theta$ curve reappears later as the energy of a current loop's magnetic moment in a magnetic field, or work through a third example where the dipole starts at $180°$ instead of $0°$?