understand how the concept of electric potential leads to the electric potential energy of two point charges and use EP = Qq / (4πε0 r)

Electric Potential – Cambridge IGCSE/A‑Level (9702)

1. Definition of electric potential

  • Concept: The electric potential \(V\) at a point is the work done by an external agent in bringing a unit positive test charge from a reference point to that point, without acceleration.
  • Reference point: By convention the reference is taken at infinity (\(r\rightarrow\infty\)), where the field of an isolated point charge is zero, giving a convenient zero‑potential baseline.


    If another reference (ground, an equipotential surface, etc.) is used it must be stated explicitly and the potential is measured relative to that surface.

  • Mathematically

    \[

    V = \frac{W{\infty\to P}}{q{\text{test}}}\qquad\bigl[{\rm V}= {\rm J\,C^{-1}}\bigr],

    \]

    where \(W_{\infty\to P}\) is the external work required to move the test charge from infinity to point \(P\).

2. Potential of a single point charge

For a point charge \(Q\) the electric field is

\[

\mathbf{E}(r)=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r^{2}}\hat{r}.

\]

The work to bring a unit test charge from \(\infty\) to a distance \(r\) is

\[

W{\infty\to r}= -\int{\infty}^{r}\mathbf{E}\!\cdot d\mathbf{s}

= -\int{\infty}^{r}\frac{1}{4\pi\varepsilon{0}}\frac{Q}{s^{2}}\,ds

= \frac{Q}{4\pi\varepsilon_{0}r}.

\]

Hence the electric potential due to a single point charge is

\[

\boxed{V(r)=\frac{Q}{4\pi\varepsilon_{0}r}=k\frac{Q}{r}}\qquad\bigl[{\rm V}= {\rm J\,C^{-1}}\bigr],

\]

with \(k=1/(4\pi\varepsilon_{0})\approx 8.99\times10^{9}\;{\rm N\,m^{2}\,C^{-2}}\).

  • Positive \(Q\) gives a positive potential, negative \(Q\) gives a negative potential.

3. Superposition – potential of a system of point charges

Because electric potential is a scalar, the total potential at a point is the algebraic sum of the contributions from each charge:

\[

\boxed{V{\text{total}} = \sum{i}\frac{k\,Q{i}}{r{i}}}

\]

where \(r{i}\) is the distance from charge \(Q{i}\) to the point of interest.

Example (two‑charge system)

\[

V(P)=k\frac{Q{1}}{r{1}}+k\frac{Q{2}}{r{2}}.

\]

4. Potential difference and equipotential surfaces

  • Potential difference between points \(A\) and \(B\):

    \[

    \Delta V = V{B}-V{A}= -\int_{A}^{B}\mathbf{E}\!\cdot d\mathbf{s}.

    \]

  • Equipotential surface: A surface on which the potential is the same everywhere (\(V=\text{constant}\)).

    • Equipotentials are always perpendicular to electric field lines.
    • For a single point charge they are concentric spheres centred on the charge.

5. Relation between electric field and potential

  • General vector form: \(\displaystyle \mathbf{E}= -\nabla V\).
  • One‑dimensional form: \(\displaystyle E_{x}= -\frac{dV}{dx}\).
  • Uniform field example:

    For a constant field \(\mathbf{E}=E\,\hat{x}\),

    \[

    V(x)= -Ex + C,

    \]

    so the potential falls linearly with distance. The constant \(C\) is set by the chosen reference (often \(V=0\) at \(x=0\)).

6. Electric potential energy

6.1 Two point charges

If a charge \(q\) is placed in the potential \(V\) created by another charge \(Q\), the electric potential energy of the pair is

\[

U = qV = \frac{k\,Qq}{r}= \frac{Qq}{4\pi\varepsilon_{0}r}.

\]

6.2 System of many point charges

For \(n\) point charges the total electrostatic potential energy is

\[

\boxed{U = \frac{1}{2}\sum{i=1}^{n}\sum{\substack{j=1\\j\neq i}}^{n}\frac{k\,Q{i}Q{j}}{r_{ij}}}

= \frac{1}{2}\sum{i}\sum{j\neq i}\frac{Q{i}Q{j}}{4\pi\varepsilon{0}r{ij}}.

\]

The factor \(\tfrac12\) avoids double‑counting each pair \((i,j)\).

6.3 Sign convention

  • Like charges (both + or both –): \(U>0\); energy must be supplied to bring them together.
  • Opposite charges: \(U<0\); the system releases energy as the charges attract.

7. Worked examples

Example 1 – Two charges

Two point charges \(Q=+5.0\;\mu\text{C}\) and \(q=-2.0\;\mu\text{C}\) are 0.10 m apart. Find \(U\).

  1. Convert: \(Q=5.0\times10^{-6}\,\text{C},\; q=-2.0\times10^{-6}\,\text{C}\).
  2. Apply \(U=\dfrac{kQq}{r}\) with \(k=8.99\times10^{9}\,\text{N m}^{2}\text{C}^{-2}\):

    \[

    U=\frac{(8.99\times10^{9})(5.0\times10^{-6})(-2.0\times10^{-6})}{0.10}

    =-0.090\;\text{J}.

    \]

  3. Negative sign → the system is at lower energy than the separated charges; energy would be released if they moved together.

Example 2 – Three‑charge system

Charges: \(Q{1}=+3\;\mu\text{C}\) at the origin, \(Q{2}=+2\;\mu\text{C}\) at \((0,0.04\text{ m})\), \(Q_{3}=-1\;\mu\text{C}\) at \((0,0.06\text{ m})\). Find the total potential energy.

  1. Compute pair separations:

    \(r{12}=0.04\;\text{m},\; r{13}=0.06\;\text{m},\; r_{23}=0.02\;\text{m}\).

  2. Use the pair‑wise formula and include the \(\tfrac12\) factor:

    \[

    U=\frac{k}{2}\!\left(\frac{Q{1}Q{2}}{r{12}}+\frac{Q{1}Q{3}}{r{13}}+\frac{Q{2}Q{3}}{r_{23}}\right).

    \]

  3. Insert numbers (convert µC to C) and evaluate:

    \[

    U\approx\frac{8.99\times10^{9}}{2}\Bigl[\frac{(3\times10^{-6})(2\times10^{-6})}{0.04}

    +\frac{(3\times10^{-6})(-1\times10^{-6})}{0.06}

    +\frac{(2\times10^{-6})(-1\times10^{-6})}{0.02}\Bigr]

    \approx -0.045\;\text{J}.

    \]

8. Summary of key formulas

QuantitySymbolFormulaUnits
Electric potential (point charge)V\(V = \dfrac{kQ}{r}= \dfrac{Q}{4\pi\varepsilon_{0}r}\)volt (V)
Potential of many point chargesV\(V = \displaystyle\sum{i}\frac{kQ{i}}{r_{i}}\)volt (V)
Potential difference\(\Delta V\)\(\Delta V = -\displaystyle\int_{A}^{B}\mathbf{E}\!\cdot d\mathbf{s}\)volt (V)
Electric field from potential\(\mathbf{E}\)\(\mathbf{E}= -\nabla V\) (or \(E_{x}= -\dfrac{dV}{dx}\) in 1‑D)newton per coulomb (N C⁻¹)
Potential energy (two charges)U\(U = \dfrac{kQq}{r}= \dfrac{Qq}{4\pi\varepsilon_{0}r}\)joule (J)
Potential energy (many charges)U\(U = \dfrac{1}{2}\displaystyle\sum{i}\sum{j\neq i}\frac{kQ{i}Q{j}}{r_{ij}}\)joule (J)
Permittivity of free space\(\varepsilon_{0}\)\(8.854\times10^{-12}\;\text{C}^{2}\text{N}^{-1}\text{m}^{-2}\)farad per metre (F m⁻¹)

9. Common pitfalls

  • Confusing \(V\) and \(U\): \(V\) is energy per unit charge; \(U = qV\) is the total energy of the pair (or system).
  • Ignoring sign of charges: The sign determines whether the potential (and energy) is positive or negative.
  • Wrong reference point: Always state whether infinity, ground, or another equipotential is used.
  • Double‑counting in many‑charge systems: Remember the factor \(\tfrac12\) in the general energy formula.
  • Applying point‑charge formulas to extended objects: Use centre‑to‑centre distance only for spherical conductors; otherwise integrate the contributions.

10. Practical tip (Paper 5 / Paper 3)

In the laboratory you can measure a potential difference with a voltmeter. For a uniform field between parallel plates, the measured \(\Delta V\) between two points a distance \(d\) apart should satisfy \(\Delta V = Ed\). This provides a quick check of both the field strength and the calibration of the instrument.

11. Practice questions

  1. Two identical positive charges of \(3.0\;\mu\text{C}\) are placed \(0.05\;\text{m}\) apart. Calculate the electric potential energy of the system.
  2. A charge \(q = +1.5\;\mu\text{C}\) is moved from a point where the potential due to a fixed charge \(Q\) is \(200\;\text{V}\) to a point where the potential is \(50\;\text{V}\). Determine the change in potential energy of \(q\).
  3. Show, using \(U = \dfrac{kQq}{r}\), why the potential energy of two like charges increases as the separation \(r\) decreases.
  4. Derive the expression \(V = \dfrac{kQ}{r}\) starting from the definition \(V = -\displaystyle\int_{\infty}^{r}\mathbf{E}\!\cdot d\mathbf{s}\) for a radial field.
  5. For a uniform electric field of magnitude \(E = 5.0\times10^{3}\;\text{V m}^{-1}\) directed along the +x axis, find the potential difference between points A (\(x=0\)) and B (\(x=0.10\;\text{m}\)).
  6. Three charges are placed at the vertices of an equilateral triangle of side \(0.08\;\text{m}\): \(Q{1}=+2\;\mu\text{C}\), \(Q{2}=+2\;\mu\text{C}\), \(Q_{3}=-1\;\mu\text{C}\). Compute the total electrostatic potential energy of the configuration.

12. Suggested diagram

Two point charges \(Q\) (positive) and \(q\) (negative) separated by a distance \(r\). Field lines radiate from \(Q\) and terminate on \(q\); equipotential surfaces are shown as concentric spheres centred on each charge, illustrating that they are perpendicular to the field lines.