define electric potential at a point as the work done per unit positive charge in bringing a small test charge from infinity to the point

Electric Potential

Objective

To define electric potential at a point as the work done per unit positive test charge in moving the charge from a chosen reference point (usually infinity) to that point, and to develop the connections required by Cambridge International AS & A‑Level Physics (9702) – 18.5, 19.1‑19.3.

1. Definition and Reference Point

• The electric potential \$V\$ at a point \$P\$ is

  \$V(P)=\frac{W_{\text{ref}\rightarrow P}}{q}\,,\$

  where \$W_{\text{ref}\rightarrow P}\$ is the work required to bring a small positive test charge \$q\$ from the reference point to \$P\$.

• For an isolated charge distribution the reference is taken at infinity, so \$V(\infty)=0\$.
• If another reference (earth, a conducting plate, etc.) is used, the same definition holds but the numerical value of \$V\$ is shifted by a constant.

2. Potential Difference (Voltage) and the Electric Field

1. Force on the test charge: \$\mathbf{F}=q\mathbf{E}\$.
2. Infinitesimal work done *against* the field: \$dW=-\mathbf{F}\!\cdot\!d\mathbf{s}=-q\mathbf{E}\!\cdot\!d\mathbf{s}\$.
3. Dividing by \$q\$ gives the infinitesimal change in potential

   \$dV=-\mathbf{E}\!\cdot\!d\mathbf{s}\,.\$

4. Integrating between two points \$A\$ and \$B\$ (electrostatic fields are conservative) yields the potential difference

   \$\Delta V{AB}=VB-VA=-\int{A}^{B}\mathbf{E}\!\cdot\!d\mathbf{s}\,.\$

   The integral is path‑independent; only the end points matter.

3. Uniform Electric Field

• For a constant field \$\mathbf{E}=E\,\hat{x}\$ the line integral is trivial:

  \$\Delta V = -E\,\Delta x \qquad\text{or}\qquad V(x)=V_0-Ex.\$

• Consequences:

  • Equipotential planes are perpendicular to \$\mathbf{E}\$.
  • The magnitude of the field is \$E=|\Delta V|/d\$ where \$d\$ is the separation of two parallel equipotentials.

4. Potential of a Single Point Charge

1. Electric field:

   \$\mathbf{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^{2}}\hat{r}.\$

2. Using \$V=-\int_{\infty}^{r}\mathbf{E}\!\cdot\!d\mathbf{s}\$,

   \$V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}=k\frac{Q}{r}.\$

3. Sign convention: \$V>0\$ for \$Q>0\$, \$V<0\$ for \$Q<0\$; the potential falls off as \$1/r\$.

5. Conducting Sphere (or isolated conductor)

• All points on the surface of a conductor in electrostatic equilibrium are at the same potential.
• For a sphere of radius \$R\$ carrying charge \$Q\$:

  \$V{\text{surface}}=\frac{1}{4\pi\varepsilon0}\frac{Q}{R}=k\frac{Q}{R}.\$

• Inside the sphere the electric field is zero, so the potential is constant and equal to the surface value.

6. Capacitor

• For a parallel‑plate capacitor of capacitance \$C\$ holding charge \$Q\$, the potential difference between the plates is

  \$V=\frac{Q}{C}.\$

• Energy stored:

  \$U_{\text{cap}}=\frac{1}{2}CV^{2}=\frac{Q^{2}}{2C}.\$

• This relation links the potential‑energy topic (19.2) with the definition of \$V\$.

7. Electric Dipole (optional A‑Level extension)

• A dipole consists of charges \$+q\$ and \$-q\$ separated by distance \$d\$. The potential at a point \$\mathbf{r}\$ (with \$r\gg d\$) is approximated by

  \$V(\mathbf{r})\approx\frac{1}{4\pi\varepsilon_0}\frac{\mathbf{p}\!\cdot\!\hat{r}}{r^{2}},\qquad\mathbf{p}=q\mathbf{d}\$

  where \$\mathbf{p}\$ is the dipole moment.

• Although not examined in depth, the expression illustrates the scalar nature of potential and the \$1/r^{2}\$ fall‑off for dipoles.

8. Super‑position of Potentials

Because potential is a scalar, the total potential at a point due to several point charges \$Q_i\$ is the algebraic sum of the individual contributions:

\$V{\text{total}}(P)=\sum{i}\frac{1}{4\pi\varepsilon0}\frac{Qi}{r_i}\,,\$

where \$r_i\$ is the distance from \$P\$ to the \$i^{\text{th}}\$ charge. This follows directly from the linearity of the integral in Section 2.

9. Equipotential Surfaces

• An equipotential surface is a set of points that all have the same potential \$V\$.
• Since \$dV=0\$ on such a surface, \$\mathbf{E}\!\cdot\!d\mathbf{s}=0\$; therefore \$\mathbf{E}\$ is everywhere perpendicular to the surface.
• Examples:

  • Concentric spheres around a point charge.
  • Parallel planes in a uniform field.
  • The surface of a conductor in electrostatic equilibrium.

10. Potential in Simple Circuits – Potential Divider

For two series resistors \$R1\$ and \$R2\$ across a battery of emf \$V{\text{in}}\$, the voltage across \$R2\$ is

\$V{\text{out}}=V{\text{in}}\frac{R2}{R1+R_2}\,,\$

a result that is frequently required in Paper 10.3 circuit questions. The same reasoning applies to any series network of impedances (A‑Level extension).

11. Units and Dimensions

• SI unit: volt (V), where \$1\ \text{V}=1\ \text{J C}^{-1}\$.
• Dimensional formula: \$[V]=\dfrac{ML^{2}}{T^{3}I}\$ (with \$I\$ the electric current).

12. Relationship to Electric Potential Energy

The electric potential energy \$U\$ of a charge \$q\$ placed at a point of potential \$V\$ is

\$U = qV\,.\$

Consequences:

• Work done *by* the field when the charge moves from \$A\$ to \$B\$: \$W_{\text{by}}=-q\Delta V\$.
• Work required *against* the field: \$W_{\text{against}}=+q\Delta V\$.

13. Example Calculations

13.1 Potential of a Point Charge (numerical)

Find \$V\$ at \$r=0.10\,\$m from a charge \$Q=+5\;\mu\text{C}\$.

\$\$V = k\frac{Q}{r}

= \frac{9.0\times10^{9}\,\text{N m}^2\!\!/\!\text{C}^2 \times 5\times10^{-6}\,\text{C}}{0.10\;\text{m}}

= 4.5\times10^{5}\ \text{V}.\$\$

13.2 Super‑position (two charges)

Two equal charges \$+2\;\mu\text{C}\$ at \$(0,0)\$ and \$(0,0.20\,\$m\$)\$. Potential at the midpoint \$(0,0.10\,\$m\$)\$:

\$\$V = k\frac{Q}{r1}+k\frac{Q}{r2}

= \frac{9.0\times10^{9}\times2\times10^{-6}}{0.10}

+\frac{9.0\times10^{9}\times2\times10^{-6}}{0.10}

= 3.6\times10^{5}\ \text{V}.\$\$

13.3 Uniform Field

A uniform field of magnitude \$E=500\;\text{V m}^{-1}\$ exists between two parallel plates 0.04 m apart. The potential difference is

\$\Delta V = E d = 500 \times 0.04 = 20\ \text{V}.\$

13.4 Capacitor

A parallel‑plate capacitor of capacitance \$C=10\;\mu\text{F}\$ stores a charge \$Q=2\;\text{mC}\$. Find the voltage across the plates and the stored energy.

\$V = \frac{Q}{C}= \frac{2\times10^{-3}}{10\times10^{-6}} = 200\ \text{V},\$

\$U = \frac{1}{2}CV^{2}= \frac{1}{2}\times10\times10^{-6}\times(200)^{2}=0.20\ \text{J}.\$

14. Table of Common Symbols

15. Suggested Diagrams (for classroom use)

• Radial field lines from a point charge \$Q\$ with a test charge \$q\$ moving from infinity to a distance \$r\$; the integration path is shown as a dashed arrow and the equipotential sphere \$V=kQ/r\$ is labelled.
• Uniform field between two parallel plates: equipotential planes, direction of \$\mathbf{E}\$, and the relation \$V=Ed\$.
• Conducting sphere: field lines outside, zero field inside, and constant potential on the surface.
• Parallel‑plate capacitor: charge \$+Q\$ and \$-Q\$ on the plates, voltage \$V\$, and stored energy illustration.

16. Key Points to Remember

• Electric potential is a scalar; potentials add algebraically (super‑position).
• Zero potential is a matter of choice – infinity for isolated charges, Earth or any convenient point for other problems.
• In electrostatics the field is conservative, so \$\displaystyle\int\mathbf{E}\!\cdot\!d\mathbf{s}\$ depends only on the end points.
• Work done by the field: \$W{\text{by}}=-q\Delta V\$; work done against the field: \$W{\text{against}}=+q\Delta V\$.
• Equipotential surfaces are everywhere perpendicular to \$\mathbf{E}\$.
• For a uniform field \$V=-Ex\$; for a point charge \$V=kQ/r\$; for a conducting sphere \$V=kQ/R\$; for a capacitor \$V=Q/C\$.
• The potential‑divider formula follows directly from the definition of voltage in series circuits.
• Remember the sign convention: a positive source charge gives a positive potential, a negative source charge gives a negative potential.