use V = Q / (4πε0r) for the electric potential in the field due to a point charge

Electric Potential – Point Charge (Cambridge International AS & A Level Physics 9702 – Topic 18.5)

Scope and Syllabus Mapping

This note covers the entire Cambridge requirement for Electric Potential at a point and the use of the formula V = Q/(4πϵ₀r). It links directly to later parts of the syllabus:

• Electric potential energy – U = qV (topics 19.1–19.3).
• Superposition of potentials – essential for multi‑charge problems, capacitors and field calculations.
• Capacitance of an isolated sphere – C = Q/V = 4πϵ₀R (topic 20.1).
• Work done by electric fields – W = qΔV.

Formal Definition of Electric Potential

The electric potential V at a point is the work done per unit positive test charge in bringing the charge from infinity (where the potential is defined as zero) to that point:

\[ V = \frac{W_{\infty\to P}}{q}\qquad\text{(units: V = J C⁻¹)} \]

Key Formulae

Derivation of the Potential from the Electric Field

1. Radial electric field of a point charge Q: \[ \mathbf{E}(r)=\frac{Q}{4\pi\varepsilon_{0}r^{2}}\hat{r} \]
2. Potential difference between two points A and B is the negative line integral of the field: \[ V_{B}-V_{A}=-\int_{A}^{B}\mathbf{E}\!\cdot\!d\mathbf{l} \]
3. Choose A at infinity where the potential is defined as zero (\(V_{\infty}=0\)). With \(d\mathbf{l}=dr\,\hat{r}\) and \(\mathbf{E}\parallel d\mathbf{l}\): \[ V(r) = -\int_{\infty}^{r}\frac{Q}{4\pi\varepsilon_{0}r^{2}}\,dr = \frac{Q}{4\pi\varepsilon_{0}r} \]

Using the Formula – Step‑by‑Step Procedure

1. Identify the charge \(Q\) (include its sign; a positive charge gives a positive potential, a negative charge a negative potential).
2. Measure the distance \(r\) from the centre of the charge to the point of interest. Convert to metres if necessary.
3. Substitute into the appropriate form: \[ V = \frac{Q}{4\pi\varepsilon_{0}r}\quad\text{or}\quad V = k\frac{Q}{r} \]
4. Check units: \(Q\) in coulombs (C), \(r\) in metres (m); the result is in volts (V).
5. Validate the sign and magnitude: a 1 C charge at 1 m gives \(\approx 9.0\times10^{9}\) V. Use this “sanity check’’ for quick error spotting.

Superposition of Potentials

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

\[ V_{\text{total}} = \sum_{i}\frac{Q_{i}}{4\pi\varepsilon_{0}r_{i}} = \sum_{i}k\frac{Q_{i}}{r_{i}} . \]

This principle underpins most A‑Level questions involving two or more charges, parallel‑plate capacitors, and equipotential surface sketches.

Link to Other Syllabus Areas

• Electric potential energy: \(U = qV\). Knowing the potential at a point immediately gives the energy of any test charge placed there.
• Capacitance of a sphere: For an isolated sphere of radius \(R\), the surface charge behaves like a point charge at its centre, so \(V = kQ/R\) and \(C = Q/V = 4\pi\varepsilon_{0}R\).
• Work done by the field: Moving a charge \(q\) through a potential difference \(\Delta V\) requires work \(W = q\Delta V\). This connects the point‑charge potential to circuit concepts (topic 20).

Worked Example 1

Problem: A point charge of \(+5.0\;\mu\text{C}\) is placed at the origin. Find the electric potential 0.20 m from the charge.

Solution:

\[ \begin{aligned} Q &= +5.0\times10^{-6}\ \text{C},\\ r &= 0.20\ \text{m},\\[4pt] V &= k\frac{Q}{r} = (8.99\times10^{9})\frac{5.0\times10^{-6}}{0.20}\\[4pt] &= 2.25\times10^{5}\ \text{V}. \end{aligned} \] Rounded to two significant figures, \(V = +2.3\times10^{5}\ \text{V}\).

Worked Example 2 – Superposition

Problem: Two point charges are placed on the x‑axis: \(+4.0\;\mu\text{C}\) at \(x=0\) and \(-3.0\;\mu\text{C}\) at \(x=0.30\ \text{m}\). Find the potential at a point P located 0.15 m to the right of the first charge (i.e. at \(x=0.15\ \text{m}\)).

Solution:

\[ \begin{aligned} r_{1} &= 0.15\ \text{m}, & V_{1} &= k\frac{+4.0\times10^{-6}}{0.15} = 2.40\times10^{5}\ \text{V},\\[4pt] r_{2} &= 0.30-0.15 = 0.15\ \text{m}, & V_{2} &= k\frac{-3.0\times10^{-6}}{0.15} = -1.80\times10^{5}\ \text{V},\\[4pt] V_{\text{total}} &= V_{1}+V_{2} = (2.40-1.80)\times10^{5} = 6.0\times10^{4}\ \text{V}. \end{aligned} \] Thus the net potential at P is \(+6.0\times10^{4}\) V.

Common Pitfalls

• Forgetting the sign of the source charge – a negative charge yields a negative potential.
• Using centimetres or millimetres for \(r\) without converting to metres.
• Applying the point‑charge formula inside a charged conducting sphere; it is valid only for points **outside** the sphere.
• Mixing up electric field (V m⁻¹) with electric potential (V).
• When using superposition, treating the contributions as vectors; potentials add algebraically because they are scalars.

Summary Table

Quantity	Symbol	Expression	Units
Permittivity of free space	\(\varepsilon_{0}\)	\(8.854\times10^{-12}\)	C² N⁻¹ m⁻²
Coulomb constant	k	\(\displaystyle\frac{1}{4\pi\varepsilon_{0}}=8.99\times10^{9}\)	N m² C⁻²
Electric potential (point charge)	V	\(\displaystyle\frac{Q}{4\pi\varepsilon_{0}r}=k\frac{Q}{r}\)	V
Electric field (point charge)	E	\(\displaystyle\frac{Q}{4\pi\varepsilon_{0}r^{2}}=k\frac{Q}{r^{2}}\)	V m⁻¹
Potential energy of a test charge	U	\(U = qV\)	J
Capacitance of an isolated sphere	C	\(C = 4\pi\varepsilon_{0}R\)	F

Suggested Diagram

Practice Questions

1. Calculate the potential at a distance of 0.10 m from a charge of \(-2.0\;\mu\text{C}\).
2. A proton (\(q = +1.60\times10^{-19}\ \text{C}\)) is placed 5.0 cm from a point charge of \(+3.0\times10^{-9}\ \text{C}\). Determine the electric potential energy of the proton at that position.
3. Two identical point charges \(+Q\) are separated by 0.30 m. Find the electric potential midway between them.
4. Using superposition, compute the potential at a point that is 0.15 m from a charge \(+4.0\;\mu\text{C}\) and 0.25 m from a second charge \(-3.0\;\mu\text{C}\).
5. Show that the capacitance of an isolated spherical conductor of radius \(R\) is \(C = 4\pi\varepsilon_{0}R\) by relating the potential formula to \(C = Q/V\).

These notes provide a concise, syllabus‑aligned framework for mastering the electric potential of a point charge, its derivation, and its connections to potential energy, capacitance, and superposition. Mastery of sign conventions, unit handling, and the scalar nature of potential will enable you to solve the full range of Cambridge AS & A‑Level examination problems.