Provide the differential relation between electric field and potential.

Provide the differential relation between electric field and potential.

• A. E = ∇V
• B. E = ∂V/∂t
• C. E = V
• D. E = −∇V

Answer: (D)

The essential idea is that the electric field tells you how the electric potential changes in space, and it points toward regions where the potential decreases most rapidly. The precise relationship is that the electric field is the negative spatial gradient of the potential: E = −∇V. This means each component of the field is the negative rate of change of the potential along that coordinate, for example Ex = −∂V/∂x, Ey = −∂V/∂y, Ez = −∂V/∂z. Thinking about an intuitive check, near a positive point charge the potential is highest close to the charge and drops off with distance. The gradient ∇V points toward increasing potential (toward the charge), so the negative gradient points outward along the field lines, matching the known direction of E. The other proposed forms don’t fit. E = ∇V would point toward increasing potential, opposite to the actual field direction. E = ∂V/∂t involves a time derivative of the potential, not its spatial variation that defines the instantaneous field. E = V would equate a vector field to a scalar potential with different units and meaning, which isn’t correct. The correct relation gives E in volts per meter and encodes both direction and magnitude through the spatial variation of V.

The essential idea is that the electric field tells you how the electric potential changes in space, and it points toward regions where the potential decreases most rapidly. The precise relationship is that the electric field is the negative spatial gradient of the potential: E = −∇V. This means each component of the field is the negative rate of change of the potential along that coordinate, for example Ex = −∂V/∂x, Ey = −∂V/∂y, Ez = −∂V/∂z.

Thinking about an intuitive check, near a positive point charge the potential is highest close to the charge and drops off with distance. The gradient ∇V points toward increasing potential (toward the charge), so the negative gradient points outward along the field lines, matching the known direction of E.

The other proposed forms don’t fit. E = ∇V would point toward increasing potential, opposite to the actual field direction. E = ∂V/∂t involves a time derivative of the potential, not its spatial variation that defines the instantaneous field. E = V would equate a vector field to a scalar potential with different units and meaning, which isn’t correct. The correct relation gives E in volts per meter and encodes both direction and magnitude through the spatial variation of V.