--- title: Point-Charge Potential module: Electric Potential moduleNumber: 3 lessonNumber: 1 order: 301 summary: | The electrostatic force is conservative, so the work it does between two points depends only on the endpoints. That lets us trade the vector field for a single scalar attached to each point, the electric potential, the potential energy a unit charge would have there. We build potential from the work integral, fix the usual reference at infinity, and add point sources as scalars, $V=k\sum_i q_i/r_i$, avoiding the vector bookkeeping the field demands. Signed charges, the reference choice, equipotential motion, and far-field expansions each give an independent check on a result. topics: [Electric Potential] draft: false sources: - book: Tipler & Mosca ref: "Ch. 23 — Electric Potential; §§23-1–23-2" --- Electrostatic force is conservative. The work done by the electric force during a displacement from $a$ to $b$ is path independent: $$ W_{a\to b}=q\int_a^b\vec E\cdot \d \vec\ell=-\Delta U. $$ Potential energy per unit charge is electric potential. Thus $$ V_b-V_a=-\int_a^b\vec E\cdot \d \vec\ell,\qquad \Delta U=q\Delta V. $$ The volt is one joule per coulomb. Potential zero is a chosen reference; forces and energy changes depend on potential differences. For isolated charges, the conventional reference is $V(\infty)=0$. $$ % caption: A positive test charge is displaced upward against a uniform downward field $\vec E$. The field does negative work on it, so both the potential and the charge's potential energy increase along the displacement. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \foreach \x in {0,1.1,4.1,5.2} {\draw[->,black] (\x,1.3)--(\x,-0.15);} \filldraw[draw=black,fill=black!8] (2.6,-0.05) circle (0.16); \node at (2.6,-0.05) {$+$}; \draw[->,acc,very thick] (2.6,0.22)--(2.6,1.12); \node[acc,above] at (2.6,1.12) {displacement}; \node[black,below] at (5.2,-0.15) {$E$}; \end{tikzpicture} $$ ## Conservative work and potential difference Electrostatic force is conservative. Its circulation around a closed path vanishes: $$ \oint\vec E\cdot \d \vec\ell=0. $$ Consequently, work between two points has no path dependence. A quasistatic external agent supplies $W_{\rm ext}=\Delta U$; the field supplies $W_E=-\Delta U$. Potential is energy per unit charge, so the signed relation is $\Delta U=q\Delta V$. A voltage is not an energy until a charge has been specified. In a uniform field $\vec E=E\hat x$, integration gives $V(x)=V(0)-Ex$. A positive charge moving a distance $d$ along the field loses $qEd$ of potential energy. An electron moving through the same displacement gains that potential energy because its charge is negative. Signed equations prevent directional errors: $$ K_b-K_a=-q(V_b-V_a). $$ ## Scalar superposition The potential of point charges is $$ V(\vec r)=\frac{1}{4\pi\varepsilon_0}\sum_i\frac{q_i}{|\vec r-\vec r_i|}. $$ Source-charge signs are included algebraically. Equal and opposite charges have $V=0$ at points equidistant from both, while their electric field is generally nonzero. Equal positive charges have zero field at their midpoint but positive potential. A potential value at one point cannot determine the field there; $\vec E=-\nabla V$ requires spatial variation of $V$. For example, charges $+4.0\ \mathrm{nC}$ and $-4.0\ \mathrm{nC}$ lie $0.30\ \mathrm m$ apart. At their midpoint both distances are $0.15\ \mathrm m$, and the two terms in $V$ cancel. The field vectors point from the positive charge toward the negative charge and add. This example separates scalar potential addition from vector field addition. ## Assembly energy Bring $q_1$ from infinity first. Bringing $q_2$ requires work $kq_1q_2/r_{12}$. Bringing $q_3$ requires $$ W_3=kq_3\left(\frac{q_1}{r_{13}}+\frac{q_2}{r_{23}}\right). $$ The total gives the pairwise form already stated. Every distinct pair appears once. In $\tfrac12\sum_iq_iV_i$, each pair occurs twice and the factor one-half corrects that duplication. The potential $V_i$ excludes charge $i$ itself; a classical point charge has divergent self-potential, which is not part of this assembly calculation. For three charges at an equilateral triangle of side $a$, $$ U=\frac{1}{4\pi\varepsilon_0a}(q_1q_2+q_1q_3+q_2q_3). $$ Negative $U$ means the assembly releases energy overall; it does not mean that each pair is attractive. The result includes all pair signs. **Electron-volts and reference levels.** An electron-volt is $e$ times one volt: $$ 1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm J. $$ An electron accelerated through $2.0\ \mathrm{kV}$ gains $2.0\ \mathrm{keV}$ of kinetic energy when other forces are negligible. Nonrelativistically, $\tfrac12m_ev^2=e|\Delta V|$; sufficiently high voltages require relativistic kinematics. Adding a constant to all potentials changes neither field nor potential difference. For localized charge distributions, $V(\infty)=0$ is convenient. For an infinite line, potential relative to infinity diverges, but a finite difference is valid: $$ V(r_b)-V(r_a)=-\frac{\lambda}{2\pi\varepsilon_0}\ln\frac{r_b}{r_a}. $$ A finite reference radius sets the potential zero; measurable energy differences remain unchanged. **Checks.** - $kq/r$ has unit $\mathrm V$. - Potential is scalar; fields and forces require vector addition. - Source sign enters $V$; test-charge sign enters $U=qV$. - Each pair is counted once in electrostatic assembly energy. - A specified reference is required for an absolute potential. ## Equipotential interpretation Points with the same potential form an equipotential surface. Moving a charge along such a surface requires no electric work because $\Delta V=0$. The field is normal to equipotential surfaces: if it had a tangential component, a tangential displacement would change potential. Equipotential geometry fixes the local field direction. Concentric spheres are equipotentials of one point charge; planes normal to a uniform field are equipotentials of that field. The spacing of adjacent equipotential surfaces indicates field magnitude. Closely spaced surfaces represent a large potential change over a small distance and hence a large $|\vec E|$. Equal spacing in a uniform-field drawing represents constant magnitude. Equipotential surfaces never cross, since a spatial point cannot be assigned two potential values. A dipole has a zero-potential plane perpendicular to its axis through its midpoint. Field lines cross this plane normally near the axis but curve farther away. The plane is not a zero-field surface: a zero value of the scalar $V$ does not force its derivative to vanish. **Potential versus force calculations.** Potential is often the economical route for a scalar energy question. To obtain a force from potential energy, differentiate with respect to the relevant coordinate: $$ F_x=-\frac{\d U}{\d x}=-q\frac{\d V}{\d x}=qE_x. $$ A charge constrained to a line has a one-dimensional derivative containing the complete force along that line. A stationary point of $U$ has zero constrained force, and the sign of the second derivative classifies stable or unstable equilibrium along that coordinate. In unrestricted three-dimensional electrostatics, the potential cannot have a stable local extremum in empty space. > **Worked example.** A $3.0\ \mathrm{nC}$ point charge sits at the origin. Find the > potential $0.20\ \mathrm m$ away, and state how it changes when the distance > doubles. > > With the reference at infinity, > > $$ > V=\frac{kq}{r}=\frac{(8.99\times10^9)(3.0\times10^{-9})}{0.20}=135\ \mathrm V. > $$ > > Doubling the distance to $0.40\ \mathrm m$ halves the potential to $67\ \mathrm V$, > because $V\propto r^{-1}$. The field at the same point drops by a factor of four, > since $E\propto r^{-2}$. The two dependences are the check: a point-charge > "potential" carrying $r^2$ in its denominator has confused the field with the > potential. ## Potential of a dipole For charges $+q$ and $-q$ separated by vector $\vec d$ directed from negative to positive, the exact potential is the difference of two point-charge terms. Far from the pair, $r\gg d$, expansion gives $$ V(\vec r)=\frac{1}{4\pi\varepsilon_0}\frac{\vec p\cdot\hat r}{r^2}, \qquad \vec p=q\vec d. $$ On the dipole axis, $V=kp/r^2$ on the positive side and $-kp/r^2$ on the negative side. In the perpendicular plane, $\vec p\cdot\hat r=0$, so potential is zero. The dipole potential falls as $r^{-2}$, one power faster than a net point charge; its field falls as $r^{-3}$. A system with zero total charge has shorter-range far-field behaviour than a charged system. The interaction energy of a fixed dipole in a uniform applied field is $$ U=-\vec p\cdot\vec E. $$ This expression follows by summing the potential energies of its two charges in the external potential. Including the dipole's own field would count its assembly energy again. A dipole aligned with the field has the lower energy; the torque is obtained from $\tau=-\d U/\d \theta=pE\sin\theta$. ## Conductors and potential An electrostatic conductor has one potential throughout its material and on its surface. If two points of a conductor had different potentials, a tangential electric field would drive its mobile charges and equilibrium would not have been reached. A conductor may carry nonzero surface charge while its interior potential is constant. The potential of an isolated conducting sphere of radius $R$ carrying charge $Q$ is $$ V(r)=\begin{cases} \dfrac{1}{4\pi\varepsilon_0}\dfrac QR,&r\le R,\\ \dfrac{1}{4\pi\varepsilon_0}\dfrac Qr,&r\ge R. \end{cases} $$ The field is zero inside, while potential is generally nonzero there. Zero field and zero potential are distinct conditions. Only potential differences have direct physical significance; grounding sets a chosen conductor's potential equal to the reference value through charge exchange with Earth. > **Worked example.** A proton starts from rest where the potential is > $+500\ \mathrm V$ and drifts, with no other forces, to a region at > $+100\ \mathrm V$. Find its kinetic-energy gain. > > The potential change is $\Delta V=100-500=-400\ \mathrm V$. Energy conservation > gives the kinetic-energy change directly from the signed charge: > > $$ > \Delta K=-q\,\Delta V=-e(-400\ \mathrm V)=+400\ \mathrm{eV} > =6.4\times10^{-17}\ \mathrm J. > $$ > > The proton gains $400\ \mathrm{eV}$: a positive charge speeds up as it falls > toward lower potential. An electron ($q=-e$) between the same two points would > instead lose $400\ \mathrm{eV}$ of kinetic energy, because its potential-energy > change carries the opposite sign. The word "drop" fixes no sign on its own; the > signed product $q\,\Delta V$ does. ## Domain of the formulas The point-charge potential is valid outside the finite extent of the source. A charged sphere may be represented by a point charge only outside a spherically symmetric distribution. Inside an extended insulator, potential is calculated from the actual charge distribution or from the field by integration. In time-dependent electromagnetic situations, an electric field may not be conservative and a single global electrostatic potential difference cannot replace the line integral of the field. Point-charge and static-potential formulas require electrostatic source configurations. **A systematic solution format.** In a discrete-charge potential problem, record source positions, draw the distance from each source to the stated observation point, and write one signed term $kq_i/r_i$ per source. Combine those terms before multiplying by any test charge. For work, calculate $q(V_b-V_a)$. This cancels an arbitrary reference and often simplifies the arithmetic. When a result is positive, state what quantity is positive. Positive potential, positive potential energy, and positive work by an external agent have distinct meanings. For instance, a negative test charge at positive potential has negative potential energy. Its electric force points toward decreasing potential energy, which for that charge is toward higher potential. Explicit signs remove apparent contradictions in such statements. Far-field behavior also checks a system of several charges. At distances large compared with source separations, replace all $r_i$ by the common leading distance $r$. The leading potential becomes $k(\sum_iq_i)/r$. If net charge is zero, that term must cancel and the next multipole term has faster decay. When source symmetry makes distances equal, a ring or spherical shell contributes one common distance factor to the integral. When directions differ but distances are simple, potential avoids the component bookkeeping required for the field. The field can subsequently be recovered by differentiation if a local force is needed. All distance variables must be measured from the source charge to the observation point, not from an arbitrary coordinate origin unless the source lies there. For multiple sources, distances are evaluated independently before the signed terms are added; a shared diagram prevents accidental reuse of one separation. For one point charge $q$, integration of $E=kq/r^2$ gives $$ V(r)=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}. $$ Potential is scalar, so a system of point charges gives $V=\sum_i kq_i/r_i$. Scalar addition avoids vector component bookkeeping for $\vec E$. A positive source has positive potential; a negative source has negative potential. The potential at a location may be zero while the field is nonzero. For example, at a point $0.20\ \mathrm m$ from $+3.0\ \mathrm{nC}$, $V=kq/r=135\ \mathrm V$. Bringing a $-2.0\ \mathrm{nC}$ charge from infinity to that point changes its potential energy by $qV=-2.7\times10^{-7}\ \mathrm J$. The negative value means the electric force performs positive work during the approach. $$ % caption: Two point sources contribute to the potential at $P$. The potentials add as signed scalars, $V=kq_1/r_1+kq_2/r_2$, while the fields at $P$ add as vectors. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \filldraw[draw=black,fill=black!8] (0,0) circle (0.22); \node at (0,0) {$+$}; \filldraw[draw=black,fill=black!8] (4,0) circle (0.22); \draw[black!70,thick] (3.85,0)--(4.15,0); \filldraw[draw=acc,fill=acc!10] (2,1.9) circle (0.07); \node[above] at (2,2.0) {$P$}; \draw[dashed,black] (0,0)--(2,1.9) node[midway,above left] {$r_1$}; \draw[dashed,black] (4,0)--(2,1.9) node[midway,above right] {$r_2$}; \node[below=5pt] at (0,0) {$q_1$}; \node[below=5pt] at (4,0) {$q_2$}; \end{tikzpicture} $$ **Potential-energy accounting.** For charges assembled from infinity, the total electrostatic energy is the sum over distinct pairs: $$ U=\sum_{ir_a$, the potential decreases outward. Selecting $V(r_b)=0$ makes $V(r_a)$ positive, but a different selected radius shifts every reported value by the same constant. The field obtained from $E_r=-\d V/\d r$ is unchanged. Coaxial conductors use the same logarithmic radial dependence between their surfaces; their finite outer conductor provides a natural reference. $$ % caption: Equipotentials (dashed circles) around a long uniformly charged line of density $\lambda$ seen end-on. The field is radial; the potential difference between radii $r_a$ and $r_b$ depends only on their ratio, so shifting the reference radius moves every value by a constant while leaving $\vec E$ unchanged. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black,dashed] (0,0) circle (0.75); \draw[black,dashed] (0,0) circle (1.35); \draw[black,dashed] (0,0) circle (1.95); \foreach \a in {30,60,120,150,210,240,300,330} {\draw[->,acc,thick] (0,0)--(\a:1.75);} \filldraw[draw=black,fill=black!8] (0,0) circle (0.10); \draw[black] (0.75,0.1)--(0.75,-0.1); \draw[black] (1.35,0.1)--(1.35,-0.1); \node[below right] at (0.78,-0.08) {$r_a$}; \node[below right] at (1.38,-0.08) {$r_b$}; \end{tikzpicture} $$ The logarithmic dependence results from integrating radial potential differences. Doubling a radius changes the potential by a fixed amount proportional to $\ln 2$, whereas doubling the radius of a point charge halves its potential when the infinity reference is used. The different functional form follows from source geometry, not from a different definition of voltage. **Potential on the axis of a charged ring.** Every axial source element of a uniformly charged ring of radius $a$ and total charge $Q$ has the same distance $r=\sqrt{a^2+z^2}$ from the observation point. Each element contributes the same distance factor, so scalar addition gives $$ V(z)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{\sqrt{a^2+z^2}}. $$ Potential uses the signed scalar contribution from every element; transverse components enter only in the electric-field calculation. Differentiating after the integration gives the axial field: $$ E_z=-\frac{\d V}{\d z} =\frac{1}{4\pi\varepsilon_0}\frac{Qz}{(a^2+z^2)^{3/2}}. $$ At the centre, $V(0)=kQ/a$ while $E_z(0)=0$. The nonzero potential reflects the work per unit charge needed to bring a test charge from the selected reference; zero field means only that the first spatial derivative vanishes at that point. For $z\gg a$, the ring behaves as a point charge: $V\simeq kQ/z$ and $E_z\simeq kQ/z^2$. The first correction depends on $a^2/z^2$, so the point-charge model is inaccurate near the physical size of the source. $$ % caption: Axial geometry of a uniformly charged ring of radius $a$. Every ring element lies the same distance $\sqrt{a^2+z^2}$ from $P$, so the scalar potential contributions share one denominator; transverse field components cancel by symmetry. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[thick] (0,0) ellipse (0.45 and 1.25); \filldraw[black] (0,0) circle (1.5pt); \draw[->,black] (-0.7,0)--(4.6,0) node[right] {$z$}; \filldraw[draw=acc,fill=acc!10] (3.7,0) circle (2.5pt); \node[below] at (3.7,-0.12) {$P$}; \draw[<->,black] (0.08,0.05)--(0.08,1.2) node[midway,right] {$a$}; \draw[black,dashed] (0,1.25)--(3.7,0) node[midway,above] {$r$}; \node[above left] at (0,1.25) {$Q$}; \end{tikzpicture} $$ Axial symmetry produces a common distance factor before the scalar integral is evaluated. Differentiating the resulting potential yields the local axial field. Starting from the field introduces transverse vector components that cancel only after integration for this observation point. **Dipole approximation and the size of its error.** The far-field dipole formula has a specified range of validity. On the positive axis of charges $-q$ and $+q$ separated by $d$, with observation coordinate $z>d/2$, the exact potential is $$ V(z)=kq\left(\frac{1}{z-d/2}-\frac{1}{z+d/2}\right) =\frac{kp}{z^2-d^2/4},\qquad p=qd. $$ For $z\gg d$, expanding the denominator gives $$ V(z)=\frac{kp}{z^2}\left(1+\frac{d^2}{4z^2}+\cdots\right). $$ The first omitted relative term is of order $d^2/(4z^2)$. At $z=5d$, it is about one percent; at $z=2d$, it is about six percent before higher terms are included. The approximate form describes remote points, not points merely outside the two charges. The approximation also fails near the plane between charges, where the two source distances cannot be replaced by a common $z$. $$ % caption: Normalized axial potential $Vz^2/kp$ of a finite dipole versus $z/d$. The exact curve approaches the far-dipole value (one) only for large $z/d$; the rise near the charges measures the failure of the $kp/z^2$ approximation close to the source. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,0)--(6.0,0) node[right] {$\frac{z}{d}$}; \draw[->,black] (0,0)--(0,3.2) node[above] {$\frac{Vz^2}{kp}$}; \draw[black,dashed] (0,1.9)--(5.7,1.9); \node[black,left] at (0,1.9) {$1$}; \draw[acc,very thick] plot[smooth] coordinates {(1.0,2.75)(1.3,2.52)(1.7,2.30)(2.2,2.13)(3.0,1.99)(4.0,1.93)(5.5,1.9)}; \node[acc,above right] at (1.35,2.55) {exact}; \node[black,above] at (4.9,1.9) {far form}; \draw[black,dotted] (2.0,0)--(2.0,2.16); \node[below] at (2.0,0) {$2$}; \end{tikzpicture} $$ The leading multipole term follows the net charge. If total charge is nonzero, the $1/r$ monopole term dominates at large distance and masks the dipole contribution. If total charge vanishes, the dipole term may be the leading one; a symmetric charge configuration can cancel it as well, leaving a still faster-decaying term. A far-field formula requires checks against both distance and the source moments that have been assumed nonzero. **Grounding, reference conductors, and charge transfer.** Grounding connects a conductor to a much larger conducting body conventionally assigned zero potential. Charges can then move until the connected conductors share one electrostatic potential. The final charge on the smaller conductor depends on nearby sources and geometry. An isolated positively charged sphere placed far from other objects may draw electrons from Earth until its net charge is approximately zero. The same sphere near a positive external source can retain an induced charge distribution while its potential remains zero relative to the grounded reference. Grounded specifies a potential boundary condition; uncharged specifies total charge. The conductor geometry and nearby sources determine the charge transfer and resulting field. An electrostatic solution with a grounded conductor uses $V=0$ on that conductor as a boundary value. The charge distribution then follows from the surrounding potential through the normal electric field at the surface. A voltage source can also hold two conductors at a specified difference while transferring charge between them. The reference potential and conductor geometry are boundary conditions. ## Worked three-source potential calculation $$ % caption: Geometry for a three-source potential at $P$. The source-to-point distances are the sides and diagonal of a 3-4-5 box; each enters the scalar sum independently, with no vector components. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \filldraw[draw=black,fill=black!8] (0,0) circle (0.16); \node[below] at (0,-0.1) {$q_1$}; \filldraw[draw=black,fill=black!8] (3.6,0) circle (0.16); \node[below] at (3.6,-0.1) {$q_2$}; \filldraw[draw=black,fill=black!8] (0,2.7) circle (0.16); \node[left] at (-0.15,2.7) {$q_3$}; \filldraw[draw=acc,fill=acc!10] (3.6,2.7) circle (0.07); \node[above] at (3.6,2.8) {$P$}; \draw[black,dashed] (0,0)--(3.6,2.7) node[midway,above left] {$r_1$}; \draw[black,dashed] (3.6,0)--(3.6,2.7) node[midway,right] {$r_2$}; \draw[black,dashed] (0,2.7)--(3.6,2.7) node[midway,above] {$r_3$}; \draw[<->,black] (0,-0.55)--(3.6,-0.55) node[midway,below] {$30$ cm}; \draw[<->,black] (-0.5,0)--(-0.5,2.7) node[midway,left] {$40$ cm}; \end{tikzpicture} $$ > **Worked example.** Three point charges lie in a plane: $q_1=+3.0\ \mathrm{nC}$ at > the origin, $q_2=-2.0\ \mathrm{nC}$ at $(30\ \mathrm{cm},0)$, and > $q_3=+1.0\ \mathrm{nC}$ at $(0,40\ \mathrm{cm})$. Find the potential at > $P=(30\ \mathrm{cm},40\ \mathrm{cm})$ and the potential energy of a > $-4.0\ \mathrm{nC}$ charge placed there. > > The distances from the three sources to $P$ are the sides and diagonal of a > $3$–$4$–$5$ box: > > $$ > r_1=0.50\ \mathrm m,\qquad r_2=0.40\ \mathrm m,\qquad r_3=0.30\ \mathrm m. > $$ > > Potential is a signed scalar sum, one $kq_i/r_i$ term per source: > > $$ > \begin{aligned} > V(P)&=k\left(\frac{3.0\times10^{-9}}{0.50} > -\frac{2.0\times10^{-9}}{0.40} > +\frac{1.0\times10^{-9}}{0.30}\right)\\ > &=53.9-45.0+30.0=39.0\ \mathrm V. > \end{aligned} > $$ > > A $-4.0\ \mathrm{nC}$ charge at $P$ then has potential energy > > $$ > U=q_tV=(-4.0\times10^{-9})(39.0)=-1.6\times10^{-7}\ \mathrm J. > $$ > > The energy is negative because a negative charge sits at positive potential. And > $V(P)>0$ says nothing about the field direction at $P$: the field needs the three > source vectors added componentwise, while the potential took only three signed > scalar terms. Compute the source potential from the sources alone, then bring in > the test charge through $U=q_tV$ or $\vec F=q_t\vec E$. **Radial and nonradial work around a point charge.** A point source $Q$ has radial electric field. A displacement tangent to a sphere centered on the source is perpendicular to $\vec E$, so it contributes no electric work. A route from radius $r_a$ to radius $r_b$ may contain arbitrary arcs, but only the radial parts contribute to the line integral. The resulting work done by the electric field on charge $q$ is $$ W_E=kQq\left(\frac{1}{r_a}-\frac{1}{r_b}\right). $$ The expression depends on endpoints through their radii, even when the physical route bends around obstacles. For positive $Q$ and positive $q$, outward motion has $r_b>r_a$ and gives positive work by the field. An external agent moving the charge quasistatically supplies the opposite work. A negative test charge reverses both energy and work signs while the source potential remains positive for positive $Q$. $$ % caption: Two routes between points $A$ and $B$ around a positive point source. Arc segments run tangent to the spherical equipotentials (dashed) and do zero work; only the change of radius sets the potential difference and the work. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \filldraw[draw=black,fill=black!8] (0,0) circle (0.2); \node[below] at (0,-0.12) {$+Q$}; \draw[black,dashed] (0,0) circle (1.05); \draw[black,dashed] (0,0) circle (1.95); \filldraw[black] (1.05,0) circle (1.8pt); \node[right] at (1.15,-0.05) {$A$}; \filldraw[black] (0,1.95) circle (1.8pt); \node[above] at (0,1.95) {$B$}; \draw[->,acc,thick] (1.05,0)--(1.9,0); \draw[->,acc,thick] (1.95,0) arc (0:90:1.95); \draw[->,black,thick] (1.05,0) arc (0:90:1.05); \draw[->,black,thick] (0,1.05)--(0,1.9); \node[black] at (-0.55,-0.55) {$r_a$}; \node[black] at (-1.05,-1.05) {$r_b$}; \end{tikzpicture} $$ The path-independence test also detects when electrostatic potential language has been applied outside its domain. A time-varying magnetic flux can produce an electric field with nonzero circulation, for which a closed loop has nonzero $\oint\vec E\cdot \d \vec\ell$. In that situation, the endpoint potential difference does not replace the full line integral. The radial point-charge result assumes stationary source charges and a conservative electrostatic field. **Conducting-sphere potential profile.** An isolated conducting sphere of radius $R$ and net charge $Q$ illustrates the difference between potential and field inside a conductor. Electrostatic equilibrium puts the excess charge on the outer surface. The exterior field has the point-charge form, while the interior field is zero. Integrating the exterior field from infinity and matching the potential continuously at the surface gives $$ V(r)= \begin{cases} kQ/R, & 0\le r\le R,\\ kQ/r, & r\ge R. \end{cases} $$ The horizontal interior segment is a nonzero constant unless $Q=0$ or a different reference has been selected. A positive test charge moved anywhere within the material has no potential-energy change because $\Delta V=0$. Moving it from the interior to infinity requires positive external work $qkQ/R$ when $Qq>0$. $$ % caption: Potential of a charged conducting sphere with $V(\infty)=0$. Inside, $V$ is the constant $kQ/R$; outside it falls as $kQ/r$, joining continuously at $R$ where the surface charge makes the slope change. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,0)--(5.9,0) node[right] {$r$}; \draw[->,black] (0,0)--(0,3.1) node[above] {$V$}; \draw[acc,very thick] (0,2.4)--(2.1,2.4); \draw[acc,very thick] plot[smooth] coordinates {(2.1,2.4)(2.7,1.86)(3.45,1.37)(4.35,1.02)(5.55,0.75)}; \draw[black,dashed] (2.1,0)--(2.1,2.4); \node[below] at (2.1,0) {$R$}; \node[acc,left] at (0,2.4) {$\frac{kQ}{R}$}; \node[acc,above] at (4.6,1.05) {$\frac{kQ}{r}$}; \end{tikzpicture} $$ Near a real conductor with a nonspherical shape, the constant-potential condition still holds throughout the connected metal, but the exterior profile need not be a function of one radial coordinate. Surface curvature and nearby conductors alter the local normal field and surface-charge density. The spherical result is a symmetry solution with a specified domain, not a generic profile for every metal object. **Potential of a finite line of charge.** A finite line source has a potential that can be integrated directly even when its field is not uniform enough for a one-step Gauss-law calculation. Take a uniformly charged line segment from $x=-a$ to $x=a$ with density $\lambda$. At a point on the perpendicular bisector a distance $b$ from the midpoint, the source element $\lambda\,\d x$ lies at distance $\sqrt{x^2+b^2}$. The potential is $$ V(b)=k\lambda\int_{-a}^{a}\frac{\d x}{\sqrt{x^2+b^2}} =2k\lambda\,\asinh\!\left(\frac{a}{b}\right). $$ The result assumes the infinity reference appropriate to a finite total charge. It diverges logarithmically as $b$ approaches zero because the ideal line places charge arbitrarily close to the observation point. A real charged wire has finite radius, and its surface or volume distribution replaces the line model inside that radius. Differentiation yields the field on the perpendicular bisector: $$ E_b=-\frac{\d V}{\d b} =\frac{2k\lambda a}{b\sqrt{a^2+b^2}}. $$ For $b\gg a$, the segment has total charge $Q=2a\lambda$ and the expression approaches $E_b=kQ/b^2$, the point-charge result. For $b\ll a$ but outside a thin wire, it approaches $2k\lambda/b$, the local long-line form. These two limits are separate approximations; a line segment of moderate aspect ratio requires the full expression. The potential falls only logarithmically over a range that is small compared with $a$, while its field has the reciprocal distance dependence. **Dipoles in nonuniform external potentials.** The energy $U=-\vec p\cdot\vec E$ describes a small permanent dipole in an external field that varies little over the charge separation. In a uniform field, the forces on the two charges have equal magnitudes and opposite directions. Their net force is zero, but their lines of action form a torque that tends to align $\vec p$ with $\vec E$. For angle $\theta$ between them, $$ U=-pE\cos\theta,\qquad \tau=pE\sin\theta. $$ In a nonuniform field, the two charges sample different field magnitudes. The opposite forces no longer cancel completely, and the leading translational force is $$ \vec F=\nabla(\vec p\cdot\vec E) $$ when the dipole moment is fixed. A dipole aligned with a field that grows toward the right is pulled toward the stronger region. The formula applies when the dipole is small compared with the field-variation length. Widely separated charges require direct summation of their separate forces. The distinction between torque and net force is essential in molecular and dielectric models. Uniform fields can orient polar molecules without translating their centers of mass. Field gradients can both orient and translate them. An induced dipole requires an additional relation between $\vec p$ and the local field. Its energy carries a factor one-half for linearly induced polarization. That factor accounts for the work required to polarize the object while the external field is applied. **Reference shifts and cancellation in computed potentials.** Potential is algebraic, so contributions of large opposite sign can leave a small result. > **Worked example.** At one point two source terms evaluate to > $V_1=+9.0\times10^4\ \mathrm V$ and $V_2=-8.999\times10^4\ \mathrm V$. Find the net > potential and check how much precision each term needs. > > $$ > V=V_1+V_2=(9.000-8.999)\times10^4\ \mathrm V=10\ \mathrm V. > $$ > > Four significant figures in each term survive as only two in the difference. Round > either term to three figures and both become $9.00\times10^4\ \mathrm V$, giving > $V=0$, a $100\%$ error. When large contributions nearly cancel, carry guard digits > through the signed sum and round only at the end. Reference shifts do not create this cancellation. Adding a constant $C$ to every potential changes $V_1$, $V_2$, and $V(P)$ according to the chosen reference, while every difference such as $V(P)-V(A)$ remains unchanged. Numerical conditioning is instead set by the physical geometry and charge signs. A potential difference between nearby points can often be computed more accurately by evaluating the change of each source term directly than by subtracting two separately rounded absolute potentials. A distant neutral pair exhibits cancellation with physical as well as numerical consequences. The leading $1/r$ terms of the two charges cancel, leaving the $1/r^2$ dipole potential. A calculation that retains a large residual monopole term far from a neutral pair has either used unequal charges, inconsistent source distances, or insufficient numerical precision. **A disciplined electrostatic-potential calculation.** A potential calculation begins with the source charges, their positions, the observation point, and the reference convention. The source configuration fixes $V$; a later test charge determines $U=qV$, work, and force. Treating a test charge as one of the sources changes signs and can double count its interaction. A discrete source set has contributions of the form $kq_i/r_i$. Distances are positive geometric lengths. Source charge signs enter the numerators, not the distances. A point located midway between equal opposite charges can have zero potential because the signed terms cancel, even though the field vectors add. The same midpoint between equal positive charges has zero field by vector symmetry but positive potential. Evaluate these two cases separately. They have different scalar and vector conditions. When a problem asks for work between two points, calculate the difference first: $$ W_{\rm ext}=q\,[V(B)-V(A)]. $$ The difference removes any shared additive reference and can cancel common source terms before numerical substitution. A source far from both endpoints contributes nearly the same potential to each, so it can have little effect on the work even when its absolute contribution to either potential is large. Field calculations have a different sensitivity because they use spatial derivatives and vector directions. The following checks expose most setup errors: - **Units:** each $kq_i/r_i$ term must have volts; multiplying its final sum by a charge gives joules. - **Far distance:** an isolated net charge gives $V\propto 1/r$; a neutral source set must lose that leading term. - **Sign:** a positive source gives positive potential under the infinity reference; a negative test charge reverses the sign of its potential energy. - **Geometry:** a spherical source can be treated as a point only outside the source; a finite line or surface requires the appropriate integral nearby. - **Domain:** electrostatic potential assumes stationary sources and a conservative electric field. Time-varying induction requires a line integral along the stated path. The numerical result must name the quantity it represents. A value of $-40\ \mathrm V$ is a potential relative to a chosen reference. A value of $-40\ \mathrm{eV}$ is an energy for a single elementary charge. A value of $-40\ \mathrm J$ describes energy only after a stated amount of charge has been involved. These units are related but are not interchangeable labels. **Model boundaries for point-charge potentials.** The expression $V=kq/r$ describes the exterior potential of an ideal point source or of a spherically symmetric source viewed from outside its radius. It does not describe the interior of a uniformly charged insulating sphere, a conductor with a cavity, or an arbitrary shaped electrode. In those cases, charge distribution and boundary conditions determine the spatial potential. Replacing every source by a point charge without checking the observation distance discards that information. A finite source also has a hierarchy of far-field descriptions. At distances much larger than its size, the total charge sets the leading $1/r$ potential. If total charge is zero, the dipole moment can supply a $1/r^2$ term. Higher source moments enter at still faster powers. The hierarchy is an approximation in the ratio of source size to observation distance. It does not identify the potential at the source surface, where the omitted terms can be comparable with the retained term. Material response sets another boundary. A specified free charge distribution in vacuum gives a potential through $\varepsilon_0$. In a dielectric, polarization creates bound charges and the relation between free charge, field, and potential depends on permittivity and geometry. A conductor reaches electrostatic equilibrium only after mobile charge has redistributed. A stated charge may be fixed on an insulator but free to move on a conductor. These choices change the source model before any potential integral is written. Electrostatic potential also omits magnetic induction. When source charges and currents vary in time, the electric field can have nonzero circulation around a closed loop. No single scalar function then represents the complete field in the same global way. A local scalar potential may be combined with an accompanying vector potential, but endpoint voltage alone no longer determines the work along an arbitrary route. Circuit electromotive force and induced voltage require that wider electromagnetic description. Relativistic and quantum limits are separate from the electrostatic source model. An electron gaining energy through a large voltage may require a relativistic kinetic-energy expression. Atomic-scale conductors can have quantized charge states and nonclassical transport, so a continuous classical charge density becomes an approximation. The potential-energy relation $U=qV$ remains the starting electrostatic coupling, while the motion and material response may require a more complete model. **Measurements, calibration, and uncertainty.** Voltage measurements compare terminals. A meter connected between an unknown conductor and a reference measures their potential difference through its own input impedance and capacitance. For a low-resistance circuit, a high input resistance usually draws negligible current. For a small isolated conductor, the meter lead can transfer enough charge or add enough capacitance to alter the quantity being measured. Electrostatic probes specify sensing distance, electrode area, input capacitance, and calibration geometry. Potential maps require spatial sampling. A grid spacing larger than the separation between nearby electrodes can miss steep gradients; a probe face averages the potential over a finite region instead of returning a mathematical point value. Near a sharp conductor tip, a millimetre placement error can correspond to a large voltage-gradient error because the equipotentials are closely spaced. The recorded position, reference conductor, source state, and probe orientation define a reproducible map. For independent source-charge and distance uncertainties, first-order propagation for a point-charge term has the scale $$ \delta V_i\simeq k\left(\frac{\delta q_i}{r_i} +\frac{|q_i|\,\delta r_i}{r_i^2}\right). $$ The expression identifies sensitive terms. A full uncertainty analysis accounts for their correlations and distributions. A nearby source can dominate the distance uncertainty because its contribution changes as $1/r_i^2$. Correlated position errors and cancellation between large signed terms require covariance information; adding independent absolute errors blindly can produce a misleading bound. A direct voltage-difference measurement may have smaller uncertainty than subtracting two separately measured absolute potentials when their reference and source errors are shared. Calibration requires a configuration with known geometry and voltage. A parallel-plate region away from edges provides an approximate constant gradient, while a conducting sphere provides a radial reference profile. Agreement at a few locations samples only those geometries; fringing, grounding paths, humidity, insulating contamination, and probe motion can introduce systematic shifts. The stated uncertainty model must include those effects before potential values are reported. **Cross-checking a potential solution.** Differentiate a derived potential whenever a corresponding field expression is available. A point charge gives $V=kq/r$ and $E_r=-\d V/\d r=kq/r^2$ with the radial direction supplied by the sign of $q$. A finite line potential differentiated with respect to perpendicular distance must approach both the point-charge and long-line limits in their stated ranges. These comparisons test the geometry, the reference, and the power of distance in one step. Energy provides an independent sign check. If a positive charge moves outward from a positive source without external support, field work is positive, potential energy decreases, and kinetic energy increases. Reversing either charge reverses the energy signs while leaving the source potential convention unchanged. State the endpoint positions and the sign of the transported charge before assigning words such as _rise_, _drop_, or _gain_ to a voltage calculation. For several sources, inspect both a nearby limit and a distant limit. Approaching one source should recover its local singular or finite-source behavior. Moving far away should recover the total charge or the first nonzero multipole moment. A result that violates either limit has lost a source term, assigned an incorrect distance, or used a model outside its stated domain. **Reference selection and physical boundaries.** Potential values require a reference, and that reference has a physical realization in an experiment. A grounded enclosure, an Earth connection, a source return lead, or a remote reference electrode acts as a charge reservoir and fixes one conductor potential. Treating that object as absent while assigning zero potential to a nearby surface changes the boundary conditions. A calculation may choose infinity as the reference only when the source arrangement and surrounding conductors make that limit meaningful. An isolated conductor differs from a grounded copy of the same shape. Its net charge is constrained, so changing the position of a nearby charge changes its potential. A grounded conductor can exchange charge with the reservoir to maintain its potential. Image-charge methods encode this distinction through their boundary condition. The force on an external charge can be found from the image construction, while the complete energy accounting must include the reservoir when charge flows to or from ground. Potential differences measured in an apparatus also depend on lead routing when fields vary in time. In the electrostatic limit, any path between two endpoints gives the same potential difference. A changing magnetic flux removes that path independence, so a pair of meter leads forms part of the electromagnetic loop. Stating the static-field approximation and the reference conductor preserves the meaning of the scalar-potential calculation.