Theory

Electric potential and electrostatic energy

Why the electrostatic field is conservative, how potential energy and potential are defined, how the potential of point charges and continuous distributions is computed, how the field is obtained from the potential and how conservation of energy describes the motion of charges.

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Work done by the electric force
Conservative electrostatic field

Work done by the electric force

Throughout the topic the source charges are at rest in an inertial reference frame, and the field they produce is called electrostatic.

A test charge $q_{0}$ moves along a curve $γ$ from $A$ to $B$ in an electrostatic field $E$ , and the force $F = q_{0} E$ acts on it. The work done by the electric force is

W_{A \to B} = \int_{γ} q_{0} E \cdot d ℓ = q_{0} \int_{γ} E \cdot d ℓ .

In this subject work is denoted by $W$ , because the letter $L$ is reserved for inductance. The work is proportional to $q_{0}$ , so the ratio $\frac{W _{A \to B}}{q _{0}}$ depends only on the field and on the curve. In a uniform field and along a segment $s$ the definition reduces to $W = q_{0} E \cdot s$ ; for example, a proton that moves $2.0 cm$ in the direction of a uniform field of $5.0 \cdot 1 0^{3} N/C$ receives the work $1.60 \cdot 1 0^{- 19} \cdot 5.0 \cdot 1 0^{3} \cdot 0.020 J = 1.6 \cdot 1 0^{- 17} J$ .

Work in the field of a point charge

Let $Q$ be a point charge at rest and let $q_{0}$ be a test charge that goes from a point $A$ , at distance $r_{A}$ from $Q$ , to a point $B$ , at distance $r_{B}$ , along any curve. The work done by the electric force is

W_{A \to B} = k Q q_{0} (\frac{1}{r _{A}} - \frac{1}{r _{B}}) .

The work depends only on the initial and final distances, so it is the same for all curves joining $A$ and $B$ . If the charges have the same sign and $q_{0}$ moves away from $Q$ , the work is positive, because the repulsive force has its radial component in the same direction as the displacement. For example, with $Q = 4.0 nC$ , $q_{0} = 1.0 nC$ , $r_{A} = 2.0 cm$ and $r_{B} = 6.0 cm$ one obtains $W = 8.99 \cdot 1 0^{9} \cdot 4.0 \cdot 1 0^{- 18} \cdot (50 - 16.7) J \approx 1.2 \cdot 1 0^{- 6} J$ .

Two curves from

A

B

in the field of

Q

: along

γ_{2}

the radial segment gives all the work and the circular arc gives none, and along

γ_{1}

the same work is obtained

Proof

The field of $Q$ is $E = \frac{k Q}{r ^{2}} \overset{r}{^}$ , as seen in the topic on the electric field. An infinitesimal displacement $d ℓ$ along the curve is decomposed into a part along $\overset{r}{^}$ , of signed length $d r$ , and a part perpendicular to $\overset{r}{^}$ . The field is radial, so the perpendicular part gives no contribution and $E \cdot d ℓ = \frac{k Q}{r ^{2}} d r$ . Therefore

W_{A \to B} = q_{0} \int_{r_{A}}^{r_{B}} \frac{k Q}{r ^{2}} d r = k Q q_{0} [- \frac{1}{r}]_{r_{A}}^{r_{B}} = k Q q_{0} (\frac{1}{r _{A}} - \frac{1}{r _{B}}) .

The result contains only $r_{A}$ and $r_{B}$ , so it does not depend on the curve.

Conservative electrostatic field

Circulation of the electrostatic field

The electrostatic field $E$ produced by any system of charges at rest is conservative: the work done by the electric force on a test charge between two points $A$ and $B$ is the same for all curves joining them. Equivalently, the circulation of $E$ along every closed curve $γ$ is zero,

\oint_{γ} E \cdot d ℓ = 0.

The electric potential energy and the potential are based on this property. A charge that travels along a closed curve and returns to its starting point receives zero total work from the electrostatic field; for this reason, a generator is needed to make a current flow in a closed circuit, as seen in the topic on electric current and direct-current circuits.

Proof

For a single point charge one applies the work in the field of a point charge. On a closed curve the initial point and the final point coincide, so $r_{A} = r_{B}$ and the work is zero. For a system of charges $Q_{1}, \dots, Q_{n}$ the field is the sum of the fields $E_{i}$ of the individual charges, by the superposition principle, and the line integral is linear, therefore

\oint_{γ} E \cdot d ℓ = i = 1 \sum n \oint_{γ} E_{i} \cdot d ℓ = 0.

For a continuous distribution the argument is the same, with an integral over the charge in place of the sum. Finally, if $γ_{1}$ and $γ_{2}$ are two curves from $A$ to $B$ , the closed curve formed by $γ_{1}$ and by $γ_{2}$ traversed backward has zero circulation; so the integrals of $E$ along $γ_{1}$ and along $γ_{2}$ are equal.

Local form of zero circulation

By Stokes’ theorem the circulation of $E$ along the boundary of a surface is the flux of the curl of $E$ through the surface. The circulation is zero along every closed curve, so the flux of the curl is zero through every surface, and therefore

\nabla \times E = 0

at every point. The relation holds for fields produced by charges at rest. In the presence of time-varying magnetic fields the circulation of the electric field is nonzero; see the topic on electromagnetic induction.

The theory continues with 8 more sections (electric potential energy, electric potential and potential difference, potential of point charges…), plus all worked examples, the solutions and the PDF.

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