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Formulas/physics/Electrostatic Potential/Electric Field from Potential (Vector)

Electric Field from Potential (Vector)

Full vector relation between field and potential. The gradient operator ∇ gives the direction of maximum rate of increase of V; the negative sign reverses it to give E.
Class 12
Derivation

Extension to three dimensions

In 3D, the work done by the field on charge q0q_0 for displacement dr⃗d\vec{r}:

dW=q0E⃗⋅dr⃗=−q0 dVdW = q_0\vec{E}\cdot d\vec{r} = -q_0\, dV

But dVdV in three dimensions is:

dV=∂V∂xdx+∂V∂ydy+∂V∂zdz=∇V⋅dr⃗dV = \frac{\partial V}{\partial x}dx + \frac{\partial V}{\partial y}dy + \frac{\partial V}{\partial z}dz = \nabla V \cdot d\vec{r}

Comparing:

E⃗⋅dr⃗=−∇V⋅dr⃗\vec{E}\cdot d\vec{r} = -\nabla V \cdot d\vec{r}

Since this holds for arbitrary dr⃗d\vec{r}:

E⃗=−∇V=−(∂V∂xi^+∂V∂yj^+∂V∂zk^)\boxed{\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)}

Verification with point charge

V=q/4πε0rV = q/4\pi\varepsilon_0 r. In spherical coordinates ∇V=(dV/dr)r^\nabla V = (dV/dr)\hat{r}:

E⃗=−dVdrr^=−ddr(q4πε0r)r^=q4πε0r2r^✓\vec{E} = -\frac{dV}{dr}\hat{r} = -\frac{d}{dr}\left(\frac{q}{4\pi\varepsilon_0 r}\right)\hat{r} = \frac{q}{4\pi\varepsilon_0 r^2}\hat{r} \checkmark
Note
The gradient operator in Cartesian coordinates is $\nabla = \partial/\partial x\,\hat{i} + \partial/\partial y\,\hat{j} + \partial/\partial z\,\hat{k}$. In spherical: $\nabla V = \partial V/\partial r\,\hat{r} + (1/r)\partial V/\partial\theta\,\hat{\theta} + \ldots$ Use the appropriate form for the symmetry of the problem.