Amit & Deepali Sinha Foundation

A Linear Potential, a Uniform Field

Irodov 3.35DifficultyUniqueness

Problem Irodov 3.35

The potential of an electrostatic field is φ=a⋅r\varphi = \vect{a}\cdot\vect{r}, where a\vect{a} is a constant vector and r\vect{r} is the radius vector of the field point. Find the electric field vector E\vect{E}.

The constant vector a and the radius vector r from the origin O to a field point P.

The constant vector a\vect{a} and the radius vector r\vect{r} from the origin OO to a field point PP.

Given
  • φ=a⋅r\varphi = \vect{a}\cdot\vect{r}
  • a\vect{a} constant
Find
the field E\vect{E}

Solution

Step 1The potential in components

With a=axi+ayj+azk\vect{a} = a_x\vect{i} + a_y\vect{j} + a_z\vect{k} and r=xi+yj+zk\vect{r} = x\vect{i} + y\vect{j} + z\vect{k}, the dot product is

φ=axx+ayy+azz(1) \varphi = a_x x + a_y y + a_z z \tag{1}
Step 2Minus the gradient

The field is minus the gradient of the potential. Each partial derivative of (1) picks out one constant component of a\vect{a}:

E=−(∂φ∂x i+∂φ∂y j+∂φ∂z k)=−(axi+ayj+azk)\begin{aligned} \vect{E} &= -\left(\frac{\partial\varphi}{\partial x}\,\vect{i} + \frac{\partial\varphi}{\partial y}\,\vect{j} + \frac{\partial\varphi}{\partial z}\,\vect{k}\right)\\ &= -\left(a_x\vect{i} + a_y\vect{j} + a_z\vect{k}\right) \end{aligned}
E=−a\vect{E} = -\vect{a}
(2)

The field is uniform: the same magnitude aa everywhere, pointing opposite to a\vect{a}.

Drawn for a at 30° above the x axis (a itself not shown): uniform field lines (crimson) running down to the left at 210°, and equipotentials (teal) at equal steps of φ crossing them at right angles, planes seen edge-on.

Drawn for a\vect{a} at 30∘30^\circ above the xx axis (a\vect{a} itself not shown): uniform field lines (crimson) running down to the left at 210∘210^\circ, and equipotentials (teal) at equal steps of φ\varphi crossing them at right angles, planes seen edge-on.

Check

The potential grows fastest along a\vect{a}, at the rate aa per unit length, and the field points down the potential, the opposite way, with exactly that magnitude. The equipotentials a⋅r=const\vect{a}\cdot\vect{r} = \text{const} are planes perpendicular to a\vect{a}, and the field lines cross them at right angles, as they must. Units: φ\varphi in volts and rr in metres make aa volts per metre, a field.

Remark

Every smooth potential looks like this one close enough to any point where the field is not zero: to first order φ≈φ0+(∇φ)⋅δr\varphi \approx \varphi_0 + (\nabla\varphi)\cdot\delta\vect{r}, so locally the field is uniform and equal to −∇φ-\nabla\varphi. A uniform field is the first term of every field, and the gradient is exactly the constant vector a\vect{a} of that local linear potential.