recall and use the fact that the electric field at a point is equal to the negative of potential gradient at that point

Electric Potential – Content 18.5 (Cambridge International AS & A Level Physics 9702)

Learning Outcome

Students must be able to recall and use the fact that the electric field at a point is equal to the negative of the potential gradient at that point:

Recall: \(\displaystyle \mathbf{E}= -\nabla V\)

Use: Given any scalar potential function \(V(x,y,z)\) (or a sum of such functions) obtain the vector electric field \(\mathbf{E}\) by differentiating and applying the minus sign.

Scope

This relationship must be applied to all configurations listed in the syllabus table (point charge, uniform field, infinite line charge, infinite sheet, and composite systems such as two‑point‑charge or dipole arrangements). The same method works when the total potential is the algebraic sum of several simpler potentials.

Key Definitions (AO1 – knowledge)

  • Electric Potential, \(V\): work done per unit positive test charge in bringing the charge from infinity to a point; \(V\) has units V = J C⁻¹.

  • Potential Energy, \(U\): energy of a charge \(q\) at a point of potential \(V\); \(U = qV\).

  • Electric Field, \(\mathbf{E}\): force per unit positive charge; units N C⁻¹ = V m⁻¹.

  • Gradient, \(\nabla\): operator that converts a scalar field into a vector field; \(\nabla V\) gives the direction and rate of greatest increase of \(V\).

  • Sign convention: field lines point from higher to lower potential, i.e. \(\mathbf{E}\) points toward decreasing \(V\).

Coordinate‑System Reminder

Use the form of \(\nabla\) that matches the symmetry of the problem.

  • Cartesian (x, y, z): \(\displaystyle \nabla =\frac{\partial}{\partial x}\hat{\mathbf i}

    +\frac{\partial}{\partial y}\hat{\mathbf j}

    +\frac{\partial}{\partial z}\hat{\mathbf k}\)

  • Spherical (r, θ, φ): \(\displaystyle \nabla =\hat{\mathbf r}\frac{\partial}{\partial r}

    +\hat{\boldsymbol\theta}\frac{1}{r}\frac{\partial}{\partial\theta}

    +\hat{\boldsymbol\phi}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}\)

  • Cylindrical (ρ, φ, z): \(\displaystyle \nabla =\hat{\boldsymbol\rho}\frac{\partial}{\partial\rho}

    +\hat{\boldsymbol\phi}\frac{1}{\rho}\frac{\partial}{\partial\phi}

    +\hat{\mathbf z}\frac{\partial}{\partial z}\)

Choose the system that makes the potential expression simplest.

Fundamental Relationships

For a single point charge the scalar potential is

\[

V(r)=\frac{Q}{4\pi\varepsilon_{0}r}=k\frac{Q}{r},

\qquad\text{where }k=\frac{1}{4\pi\varepsilon_{0}}\approx 8.99\times10^{9}\ \text{N m}^{2}\text{C}^{-2}.

\]

Vector relationship between field and potential (valid in any coordinate system):

\[

\mathbf{E}= -\nabla V

=-\left(\frac{\partial V}{\partial x}\hat{\mathbf i}

+\frac{\partial V}{\partial y}\hat{\mathbf j}

+\frac{\partial V}{\partial z}\hat{\mathbf k}\right).

\]

Derivation from Work‑Energy (AO2 – reasoning)

  1. Infinitesimal work on a charge \(q\) displaced by \(d\mathbf{s}\): \(dW = q\,\mathbf{E}\!\cdot\!d\mathbf{s}\).

  2. Corresponding change in potential energy: \(dU = q\,dV\).

  3. Conservation of energy gives \(dU = -dW\): \(q\,dV = -q\,\mathbf{E}\!\cdot\!d\mathbf{s}\) ⇒ \(dV = -\mathbf{E}\!\cdot\!d\mathbf{s}\).

  4. Dividing by the infinitesimal displacement and recognising the definition of the gradient yields the fundamental result

    5. \[

    \boxed{\mathbf{E}= -\nabla V}.

    \]

Practical Use – Step‑by‑Step Procedure (AO2)

  1. Write the scalar potential function \(V(x,y,z)\) for the configuration (state the reference where \(V=0\)).

  2. Choose the coordinate system that matches the symmetry.

  3. Calculate the gradient \(\nabla V\) using the appropriate form.

  4. Insert the minus sign: \(\mathbf{E}= -\nabla V\).

  5. Check units: \(V\) in volts, \(\nabla V\) in V m⁻¹, giving \(\mathbf{E}\) in N C⁻¹.

  6. If the total potential is a sum of several contributions, differentiate each term and add the resulting field vectors.

Common Potential Functions (and boundary‑condition note)

ConfigurationPotential \(V\) (relative to infinity)Electric Field \(\mathbf{E}= -\nabla V\)
Point charge \(Q\) at the origin\(\displaystyle V(r)=\frac{Q}{4\pi\varepsilon_{0}r}=k\frac{Q}{r}\)\(\displaystyle \mathbf{E}= \frac{Q}{4\pi\varepsilon_{0}r^{2}}\hat{r}=k\frac{Q}{r^{2}}\hat{r}\)
Uniform field (parallel plates, field along \(x\))\(\displaystyle V(x)= -E_{0}x + C\)\(\displaystyle \mathbf{E}=E_{0}\hat{\mathbf i}\) (constant, perpendicular to equipotentials)
Infinite line charge, linear density \(\lambda\)\(\displaystyle V(\rho)=\frac{\lambda}{2\pi\varepsilon{0}}\ln\!\left(\frac{\rho{0}}{\rho}\right)\) (reference radius \(\rho_{0}\))\(\displaystyle \mathbf{E}= \frac{\lambda}{2\pi\varepsilon_{0}\rho}\,\hat{\boldsymbol\rho}\)
Infinite charged sheet, surface density \(\sigma\)\(\displaystyle V(z)= -\frac{\sigma}{2\varepsilon_{0}}\,z + C\) (zero at the sheet)\(\displaystyle \mathbf{E}= \frac{\sigma}{2\varepsilon_{0}}\,\hat{\mathbf n}\) (constant, normal to the sheet)

Boundary‑condition note: Across a charged sheet the normal component of \(\mathbf{E}\) changes by \(\sigma/\varepsilon_{0}\); the potential remains continuous, which is reflected in the linear form of \(V(z)\) above.

Equipotential Surfaces

  • Surfaces on which \(V\) is constant are called equipotentials.

  • Because \(dV = -\mathbf{E}\!\cdot\!d\mathbf{s}\), any displacement \(d\mathbf{s}\) that stays on an equipotential gives \(dV=0\); therefore \(\mathbf{E}\) is always perpendicular to equipotential surfaces.

  • The magnitude of the field equals the rate of change of potential normal to the surface:

    \[

    |\mathbf{E}| = \frac{\Delta V}{\Delta s_{\perp}},

    \]

    where \(\Delta s_{\perp}\) is the perpendicular distance between adjacent equipotentials.

  • Quick‑check: From \(V(x)= -E{0}x + C\) show that \(\mathbf{E}=E{0}\hat{\mathbf i}\) and that the equipotentials are planes of constant \(x\), each perpendicular to the field.

  • Examples:

    • Point charge – concentric spherical equipotentials.

    • Uniform field – parallel planes.

    • Infinite line – cylindrical surfaces.

Worked Examples

1. Point Charge (gradient method – spherical coordinates)

Given: \(Q = +5.0\;\mu\text{C}\) at the origin. Find \(\mathbf{E}\) at \((0,0,0.10\ \text{m})\).

  1. Potential: \(V(r)=kQ/r\).

  2. In spherical symmetry \(\nabla V = \dfrac{dV}{dr}\hat{r}\).

  3. \(\displaystyle \frac{dV}{dr}= -k\frac{Q}{r^{2}}\) ⇒ \(\nabla V = -k\frac{Q}{r^{2}}\hat{r}\).

  4. \(\mathbf{E}= -\nabla V = k\frac{Q}{r^{2}}\hat{r}\).

  5. Insert numbers (\(r=0.10\ \text{m}\)):

    \[

    \mathbf{E}= \frac{8.99\times10^{9}\times5.0\times10^{-6}}{(0.10)^{2}}\;\hat{z}

    = 4.5\times10^{6}\ \text{N C}^{-1}\,\hat{z}.

    \]

2. Uniform Field from a Linear Potential

Given: \(V(x)= -E{0}x + C\) with \(E{0}=200\ \text{V m}^{-1}\).

  • \(\displaystyle \frac{dV}{dx}= -E{0}\) ⇒ \(\nabla V = -E{0}\hat{\mathbf i}\).

  • \(\mathbf{E}= -\nabla V = E_{0}\hat{\mathbf i}\) (points in the +x direction, perpendicular to the equipotential planes of constant \(x\)).

3. Infinite Line Charge (cylindrical coordinates)

Given: \(\lambda = 2.0\ \mu\text{C m}^{-1}\), reference radius \(\rho_{0}=1\ \text{m}\).

  1. Potential: \(V(\rho)=\dfrac{\lambda}{2\pi\varepsilon{0}}\ln\!\left(\dfrac{\rho{0}}{\rho}\right)\).

  2. In cylindrical coordinates \(\nabla V = \dfrac{\partial V}{\partial\rho}\hat{\boldsymbol\rho}\).

  3. \(\displaystyle \frac{\partial V}{\partial\rho}= -\frac{\lambda}{2\pi\varepsilon{0}\rho}\) ⇒ \(\nabla V = -\frac{\lambda}{2\pi\varepsilon{0}\rho}\hat{\boldsymbol\rho}\).

  4. \(\mathbf{E}= -\nabla V = \frac{\lambda}{2\pi\varepsilon_{0}\rho}\hat{\boldsymbol\rho}\).

4. Composite Potential – Two Point Charges

For charges \(Q{1}\) at \(\mathbf{r}{1}\) and \(Q{2}\) at \(\mathbf{r}{2}\), the total potential is

\[

V(\mathbf{r}) = k\frac{Q{1}}{|\mathbf{r}-\mathbf{r}{1}|}+k\frac{Q{2}}{|\mathbf{r}-\mathbf{r}{2}|}.

\]

Apply \(\mathbf{E}= -\nabla V\) term‑by‑term; the resulting field is the vector sum of the fields of each charge – a direct illustration of the superposition principle.

Common Mistakes to Avoid

  • Leaving out the minus sign – the field points toward decreasing potential.

  • Treating \(\nabla V\) as a scalar; remember it is a vector with components.

  • Using inconsistent reference points for \(V\); always state where \(V=0\) (normally at infinity).

  • Confusing the constant \(k\) with \(\frac{1}{4\pi\varepsilon_{0}}\); keep the notation consistent and note the numerical value.

  • Failing to convert charge units (µC, nC, etc.) before substitution.

  • Applying the Cartesian form of the gradient to a problem with spherical or cylindrical symmetry – choose the appropriate coordinate system.

Suggested Diagram

Equipotential surfaces (concentric spheres) around a positive point charge with radial electric‑field lines. The field lines are perpendicular to the surfaces and point outward, showing that \(\mathbf{E}\) points toward lower potential.

Summary

The scalar electric potential provides a powerful shortcut for finding electric fields. By remembering the core relationship \(\boxed{\mathbf{E}= -\nabla V}\) and applying the appropriate gradient (Cartesian, spherical or cylindrical), you can obtain \(\mathbf{E}\) for any configuration covered by the Cambridge AS & A Level syllabus – from simple point charges to composite systems. Mastery of this method is directly tested in both knowledge (AO1) and reasoning (AO2) components of the examination.