Electric Potential and Energy
A vector field at every point is a lot to track. Electrostatics offers a shortcut: a single scalar, the potential, whose hill-slope is the field itself. From it flow voltage, the conservative law behind every circuit, and the energy locked in the field.
- The work done moving a charge through a field, and how it defines potential difference.
- The potential of a point charge and, by superposition, of any distribution.
- The field–potential relation \(\vec{E} = -\nabla V\) — the gradient of Chapter 3 made physical.
- Why the electrostatic field is conservative: \(\oint\vec{E}\cdot d\vec{l}=0\), Maxwell's curl equation for statics.
- The electric dipole, its potential, and the \(1/r^2\) and \(1/r^3\) falloffs.
- The energy stored in an electrostatic field and its energy density.
Work & Potential Difference
To move a charge \(Q\) against an electric field, you must do work. The field pushes with force \(Q\vec{E}\); you push the other way. The work done by you in carrying \(Q\) from point \(A\) to point \(B\) is the line integral met in Chapter 3:
The potential difference \(V_{AB}\) is this work per unit charge — the work to move one coulomb from \(B\) to \(A\). Its unit is the volt (J/C):
Potential of a Point Charge
Taking the reference point at infinity (where \(V=0\)) and integrating the point-charge field inward gives the absolute potential. Because \(\vec{E}\) is radial and falls as \(1/r^2\), its integral gives a clean \(1/r\):
Surfaces of constant \(V\) are equipotentials — spheres for a point charge. Since no work is done moving along an equipotential, the field can have no component along it, so \(\vec{E}\) is always perpendicular to equipotential surfaces. Potential is a scalar, so unlike \(\vec{E}\) it adds by simple arithmetic, not vectors.
Potential of Distributions
Because potential superposes as a scalar sum, the potential of many charges — or a continuous distribution — is just the sum or integral of point-charge potentials. No components, no angles to resolve:
It is almost always easier to compute the scalar \(V\) and then differentiate than to integrate the vector \(\vec{E}\) directly. This "potential-first" strategy is the workhorse of electrostatic problem-solving.
The Relation E = −∇V
Potential difference is a line integral of \(\vec{E}\); reversing that relationship, the field is the negative gradient of the potential. This is the gradient of Chapter 3 given physical meaning: \(\vec{E}\) points in the direction of steepest decrease of \(V\), downhill on the potential landscape.
This single equation explains the picture above: since \(\nabla V\) is perpendicular to equipotentials (Chapter 3), so is \(\vec{E}\). It also confirms the units — \(\nabla V\) has units of volts per metre, exactly the units of \(\vec{E}\).
The Conservative Field
Because \(\vec{E}\) is a gradient, the work to move a charge between two points depends only on the endpoints, never on the path. Equivalently, the work around any closed loop is zero. This makes the electrostatic field conservative:
The point form follows from Stokes' theorem (Chapter 3) and the identity that the curl of a gradient is always zero. This is Maxwell's curl equation for statics — and it is precisely Kirchhoff's voltage law: the sum of potential rises and drops around any circuit loop is zero. The circuit law you already know is electrostatics in disguise.
The Electric Dipole
An electric dipole is two equal and opposite charges \(\pm Q\) separated by a small distance \(d\). Its dipole moment is \(\vec{p} = Q\vec{d}\), pointing from the negative to the positive charge. Far from the dipole (\(r\gg d\)), the two potentials nearly cancel, leaving a field that fades faster than a single charge's:
Electrostatic Energy
Assembling a set of charges takes work, and that work is stored as electrostatic energy. Bringing in each charge against the potential of those already placed gives:
The factor of \(\tfrac{1}{2}\) avoids double-counting each pair. Remarkably, this energy can be re-expressed as living in the field itself, spread through all of space with an energy density:
Worked Examples
Problem. Find \(V\) at \(0.3\ \text{m}\) from a \(+6\ \text{nC}\) charge.
Solution. Apply \(V = Q/4\pi\varepsilon_0 r\):
Problem. How much work moves \(+2\ \text{nC}\) from a point at \(50\ \text{V}\) to one at \(120\ \text{V}\)?
Solution. Work equals charge times potential rise, \(W = Q(V_B - V_A)\):
Problem. Given \(V = x^2 y + 2z\ \text{V}\), find \(\vec{E}\).
Solution. Take the negative gradient:
Problem. Charges \(+4\ \text{nC}\) and \(-4\ \text{nC}\) sit \(0.2\ \text{m}\) apart. Find \(V\) at the point \(0.1\ \text{m}\) from each.
Solution. Add the scalar potentials — equal magnitudes, opposite signs:
Problem. Show the work moving a charge around a closed triangle in a uniform field \(\vec{E}=E_0\,\hat{a}_x\) is zero.
Solution. A conservative field gives zero circulation:
Problem. Find the energy density where \(E = 2\times10^{5}\ \text{V/m}\) in free space.
Solution. Apply \(w_E = \tfrac{1}{2}\varepsilon_0 E^2\):
Chapter Summary
\(V_{AB}=-\int_B^A\vec{E}\cdot d\vec{l}\); work per unit charge; unit volt.
\(V=\dfrac{Q}{4\pi\varepsilon_0 r}\); a scalar, so it superposes by simple addition.
Field is steepest descent of potential; always perpendicular to equipotentials.
\(\oint\vec{E}\cdot d\vec{l}=0\), \(\nabla\times\vec{E}=0\); this is Kirchhoff's voltage law.
\(\vec{p}=Q\vec{d}\); \(V\propto 1/r^2\), \(E\propto 1/r^3\) — faster falloff than a point charge.
\(W_E=\tfrac12\int\vec{D}\cdot\vec{E}\,dv\); density \(w_E=\tfrac12\varepsilon_0 E^2\); energy lives in the field.
Problems
For each item, decide whether to work with the scalar \(V\) or the vector \(\vec{E}\) first — usually \(V\) — then convert if needed. Difficulty rises down the list.
- Find the potential \(3\ \text{m}\) from a \(-12\ \text{nC}\) point charge.
- Compute the work to move \(+5\ \text{nC}\) from \(20\ \text{V}\) to \(95\ \text{V}\).
- For \(V = 4xy - z^2\ \text{V}\), find \(\vec{E}\) at \((1,1,1)\).
- Two charges \(+Q\) sit at \((\pm a,0,0)\). Find \(V\) on the \(y\)-axis and then \(\vec{E}\) there.
- A charge \(+8\ \text{nC}\) at the origin and \(-8\ \text{nC}\) at \((0.1,0,0)\) form a dipole. Find \(V\) at \((0,0,0.2)\).
- Show that the work to move a charge between two points is independent of path for \(\vec{E}=k\,\hat{a}_x\).
- A uniformly charged ring of radius \(a\) carries total charge \(Q\). Find \(V\) on its axis, then \(\vec{E}\) by differentiation.
- Compute the total energy stored by assembling three \(+Q\) charges at the corners of an equilateral triangle of side \(a\).
- The potential in a region is \(V = 50/r\ \text{V}\) (spherical). Find \(\vec{E}\) and the charge producing it.
- Find the energy density at the surface of a \(1\ \text{cm}\)-radius sphere carrying \(10\ \text{nC}\).
- Derive \(\vec{E}=-\nabla V\) for the point-charge potential and confirm it matches Coulomb's field.
- Explain, using \(\nabla\times\vec{E}=0\), why a static electric field can never form closed loops, and contrast this with the magnetic field of Part 3.