From potential to field: the gradient

The relation $V_B - V_A = -\int_A^B \vec{E}\cdot \dd\vec{\ell}$ links the potential to the field through an integration. But the same relation can be inverted: if we know $V$ at every point, we can obtain $\vec{E}$ by performing a differentiation. In three dimensions the correct operation is the gradient, that is, the vector that collects the partial derivatives of $V$ along the three axes.

Principle — Field as the gradient of the potential

The electric field is the negative of the gradient of the potential: $\ev{\vec{E} = -\,\nabla V = -\left(\frac{\partial V}{\partial x}\,\vers{x} + \frac{\partial V}{\partial y}\,\vers{y} + \frac{\partial V}{\partial z}\,\vers{z}\right)}$

The gradient points in the direction of maximum growth of $V$; the minus sign flips it, so that $\vec{E}$ points towards decreasing values of the potential, with a strength proportional to how steep the variation is.

Key formula

In one dimension, or along the direction perpendicular to the equipotentials: $E = -\dfrac{\dd V}{\dd r}$ Practical estimate over a finite difference: $\abs{\vec{E}} \approx \dfrac{\abs{\Delta V}}{\Delta d}$

Applying the gradient formula to the cases already known returns the correct fields, confirming the consistency of the framework:

• Point charge: $V = kQ/r \Rightarrow E = -\dd V/\dd r = kQ/r^2$. ✓
• Parallel-plate capacitor: $V = -Ed + C \Rightarrow E = -\dd V/\dd d = E$. ✓
• Spherical shell (inside): $V = kQ/R = \text{const} \Rightarrow E = 0$. ✓
• Infinite plane: $V = -\dfrac{\sigma}{2\varepsilon_0}\,d \Rightarrow E = \dfrac{\sigma}{2\varepsilon_0}$. ✓

In particular, inside the conducting shell the potential is constant: its derivative is zero and hence the field is zero, in perfect agreement with Gauss’s theorem.

Summary — Typical potentials

Distribution	Potential
Point charge	$V = kQ/r$
Shell ($r>R$)	$V = kQ/r$
Shell ($r<R$)	$V = kQ/R$ (constant)
Plane (zero on the plane)	$V = -\dfrac{\sigma}{2\varepsilon_0}\,d$
Wire	$V = -\dfrac{\lambda}{2\pi\varepsilon_0}\ln r + C$

Why the minus sign?

The relation $\vec{E} = -\nabla V$ says that the field points downhill in the landscape of the potential, exactly like an object sliding down a mountain moves towards lower ground. Water flows where the elevation decreases; a positive charge spontaneously moves to where $V$ decreases. The minus sign ensures that the work done by the field when $q > 0$ moves from $A$ to $B$ is $W_{A\to B} = q(V_A - V_B) > 0$ if $V_A > V_B$: the charge “descends” the potential and the field does positive work.

Collegamenti

Argomenti: Campo elettrico e potenziale Concetti: Campo elettrico · Potenziale elettrico Metodi: Metodo del gradiente

Esercizi collegati: True or false on field and potential · Cloud-to-ground potential difference · Field lines never cross