Potential and Field Relationship

The relationship between electric field and electric potential is one of the most conceptually important results in electrostatics. Once the potential V is known as a function of position, the electric field E can be obtained without performing any integral. This is practically significant because computing potential (a scalar) is usually far simpler than computing the vector field directly. The relationship connects the two fundamental quantities of electrostatics through the mathematical operation of the gradient.

Core Concept: Gradient Operation

The electric field E is related to the electric potential V by the equation E equals negative gradient of V. The gradient of a scalar function gives a vector pointing in the direction of maximum rate of increase of that function. The negative sign means E points in the direction of maximum rate of decrease of V, that is, from high potential toward low potential. This is physically consistent with the fact that a positive charge released in a field will naturally move from high to low potential, losing potential energy.

This relationship is not merely a mathematical definition. It carries deep physical meaning. The steeper the potential gradient (the more rapidly V changes with position), the stronger the electric field at that location. In regions where V is constant (equipotential regions), the gradient is zero and therefore E is zero. This is exactly why the interior of a conductor at equilibrium, which is all at the same potential, has no electric field inside it.

Mathematical Expression

In Cartesian coordinates, the gradient of V gives the three components of E. Specifically, E_x equals negative partial derivative of V with respect to x, E_y equals negative partial V over partial y, and E_z equals negative partial V over partial z. In compact notation, E equals negative del V, where del (nabla) is the vector differential operator.

The inverse relationship, obtaining V from E, involves a line integral. V at a point equals negative integral of E dot dL from the reference to that point. These two operations, gradient and line integral, are inverses of each other for conservative fields. The existence of potential relies on the fact that the curl of E is zero for static fields (Faraday's law with no time variation), which is mathematically equivalent to E being expressible as a gradient.

For a point charge Q at the origin, V equals kQ over r, where r is the radial distance. Taking the gradient in spherical coordinates gives the radial component E_r equals negative dV/dr equals kQ over r squared. This directly yields Coulomb's law, confirming the consistency between the two descriptions.

Equipotential Surfaces in Detail

An equipotential surface is a surface on which V has the same value at every point. Moving a charge along this surface requires zero work because V does not change. Since E equals negative gradient of V and the gradient is always perpendicular to surfaces of constant V, the electric field is always perpendicular to equipotential surfaces. This orthogonality is a geometric consequence of the gradient theorem.

For a point charge, equipotential surfaces are concentric spheres. For an infinite line charge, they are coaxial cylinders. For a uniform field, they are parallel planes perpendicular to the field direction. In all cases, the field lines and equipotential surfaces form a mutually orthogonal network, and the density of equipotential lines indicates the field strength (closely spaced equipotentials mean a strong field).

Given:
Electric potential varies as V = 3x^2 + 2y in a region (V in volts, x and y in meters).
Find E at point (2, 1, 0).

Why this formula applies:
E = -grad V. Compute partial derivatives with respect to x, y, z.

Formula:
E_x = -dV/dx
E_y = -dV/dy
E_z = -dV/dz

Substitution:
dV/dx = 6x, so E_x = -6x
dV/dy = 2, so E_y = -2
dV/dz = 0, so E_z = 0

At (2, 1, 0):
E_x = -6*2 = -12 V/m
E_y = -2 V/m
E_z = 0

Final Answer: E = -12 ax - 2 ay V/m at (2,1,0)
|E| = sqrt(12^2 + 2^2) = sqrt(144 + 4) = sqrt(148) = 12.17 V/m

Exam Tip: In GATE problems with V given as a polynomial, always compute E by partial differentiation using E = -grad V. Never integrate again. Common trap: forgetting the negative sign, which reverses the direction of E. Incorrect direction means wrong force prediction on test charges.

• E equals negative gradient of V. The field points from high potential to low potential, in the direction of steepest V decrease.
• In Cartesian coordinates: E_x = -dV/dx, E_y = -dV/dy, E_z = -dV/dz.
• Where V is constant (equipotential region), grad V = 0 and therefore E = 0.
• Equipotential surfaces are always perpendicular to E field lines. This orthogonality is exact and always holds for static fields.
• The curl of E equals zero for static fields. This is the mathematical condition that guarantees a unique scalar potential V exists.

Quick Revision

• E = -grad V = -(dV/dx ax + dV/dy ay + dV/dz az).
• V = -integral(E dot dL) from reference to field point.
• Positive charge moves from high V to low V (E direction).
• Equipotential surface: V = constant, E perpendicular to it, zero work to move charge along it.
• Conductor at equilibrium: entire body at one potential, E inside = 0.
• Exam trap: Forgetting the negative sign in E = -grad V gives completely wrong E direction. Always negate the gradient.
• Dense equipotential lines indicate large |E|. Sparse lines indicate weak field.