EM-09 · Electrostatics toolkitTool equip

The Gradient

The gradient extracts, from a scalar field, the direction of steepest climb and its slope as a single vector. Turning a test direction, you see that the climb is steepest when it lines up with the gradient, and confirm that the electric field points the opposite way (E = -∇V).

Find the steepest way uphill

The tilted plane is a landscape whose height represents potential, and the bright cyan lines are equipotential lines of equal potential. Turning the test arrow with the slider shows on the indicator the uphill slope in that direction, as a percentage of the maximum. It is steepest when you align it with the gold arrow, and the slope vanishes when you follow an equipotential line.

Test direction αα = 110°

Drag to rotate; the slider changes the test direction α.

The climb in this direction

∇V = (∂V/∂x, ∂V/∂y, ∂V/∂z)

E = −∇V · the field is the gradient flipped downhill

Rate in this direction (vs maximum) · 22%

Along a contour · climb 0

The gradient, steepest ascent

A scalar field gives each point one number (EM-02·EM-08). The gradient ∇V points in the direction in which that number grows fastest, and the arrow's length is the slope in that direction. It shows, as one arrow, the steepest way up a hill, so one scalar field yields one vector field.

Observe∇V = (∂V/∂x, ∂V/∂y, ∂V/∂z)

The gradient is the vector that collects the rates of change along each axis.

The directional derivative, by dot product

The rate at which the potential changes as you move in some direction û is called the directional derivative. Its value is the dot product of the gradient and the unit vector of that direction, ∇V·û, the same principle as the dot product in EM-05. When û lines up with ∇V, the cos of the angle between them is 1, so it is a maximum; at right angles it is 0. Going the exact opposite way gives the steepest descent. So along an equipotential line the rate of change is 0, straight along the gradient it is a maximum, and the size of that maximum rate is exactly |∇V|.

ChoosedV/ds = ∇V · ?

The rate of change in one direction is the dot product of the gradient and that direction's unit vector.

Fill in(dV/ds)_max = ?

The rate of change is greatest along the gradient, and its size is |∇V|.

Perpendicular to equipotentials

The direction with a rate of change of 0 is the equipotential line, so the gradient is always perpendicular to the equipotential surface (just as seen in EM-08). It is like the contour lines of a landscape meeting the steepest climb at right angles. So a single gradient holds both a direction, the normal to the equipotential surface, and a size, the slope.

E = -∇V, from scalar to vector

The electric field points the way the potential falls fastest, so it is the exact opposite of the gradient. E = -∇V. The minus sign turns uphill into downhill. This one line draws the whole vector field, the electric field, out of the scalar field, the potential. Knowing the potential alone, you get the three components of the electric field by differentiating, and conversely, integrating the electric field along a path brings you back to the potential.

On your ownE = ?

The electric field points opposite to the potential gradient.

Back to the first screen

On the first screen, turning the test arrow made the uphill slope differ with direction. It was 0 along an equipotential line and largest when lined up with the gradient. What holds that maximum direction and size together is the gradient ∇V, and the directional derivative ∇V·û states this with a single cosine. The electric field is this gradient flipped downhill, E = -∇V, which is why it was always perpendicular to the equipotential surfaces.

The gradient ∇V extracts, from a scalar field V, the direction of steepest climb and the size of that slope as a single vector. Its components are the partial derivatives along each axis, (∂V/∂x, ∂V/∂y, ∂V/∂z). The rate of change in a direction û is the dot product ∇V·û, greatest (=|∇V|) when û lines up with ∇V. The gradient is perpendicular to equipotential surfaces, and the electric field is its opposite, E = -∇V.

Tool equipOnce you equip this tool

The gradient is a tool that extracts a vector field from a scalar field. With it equipped, you get the electric field E = -∇V straight from the potential (recovering EM-08), and in any problem with a potential you find the field by differentiating. Together with divergence ∇·(EM-07) and curl ∇×(EM-16), it makes up the three uses of the nabla operator ∇ and supports the differential form of Maxwell's equations.