recall and use the fact that the electric field at a point is equal to the negative of potential gradient at that point

Electric Potential & Electric Field

The Core Relationship

In physics, the electric field at a point is the negative gradient of the electric potential at that point:

\$\mathbf{E} = -\nabla V\$

Think of potential as the “height” of a hill and the electric field as the direction a ball would roll down. The steeper the slope (larger gradient), the stronger the field.

Gradient in Cartesian Coordinates

ComponentMathematical Form
\$E_x\$\$-\,\dfrac{\partial V}{\partial x}\$
\$E_y\$\$-\,\dfrac{\partial V}{\partial y}\$
\$E_z\$\$-\,\dfrac{\partial V}{\partial z}\$

Practical Example: Point Charge

For a point charge \$q\$ at the origin, the potential is

\$V(r) = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}\$

Take the gradient:

  1. Compute \$\dfrac{\partial V}{\partial r} = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}\$

  2. Since the field is radial, \$\mathbf{E} = -\dfrac{\partial V}{\partial r}\,\hat{r}\$

  3. Result: \$\mathbf{E}(r) = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}\,\hat{r}\$

Notice the negative sign cancels the negative derivative, giving a field that points away from a positive charge.

Exam Tip Box

⚡️ Quick Checklist for Exam Questions:

  • Always write \$\mathbf{E} = -\nabla V\$ before starting.

  • Check the sign: a decreasing potential in a direction gives a field in that direction.

  • Remember units: \$V\$ in volts (V), \$\mathbf{E}\$ in volts per metre (V/m).

  • For spherical symmetry, use \$r\$‑derivatives only.

  • Show vector direction with unit vectors (\$\hat{r}\$, \$\hat{\theta}\$, \$\hat{\phi}\$).

Analogy: Water Flow

Imagine a lake with a hill in the middle. The water level (potential) is higher near the hill. Water flows downhill, following the steepest descent—just like an electric field follows the steepest decrease in potential.

Quick Practice Problem

A uniform electric potential varies linearly with \$x\$: \$V(x) = 5x\$ V. What is the electric field?

  1. Compute \$\dfrac{dV}{dx} = 5\$ V/m.

  2. Apply \$\mathbf{E} = -\dfrac{dV}{dx}\,\hat{x}\$.

  3. Answer: \$\mathbf{E} = -5\,\hat{x}\$ V/m (points in the negative \$x\$ direction).

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Remember: a positive slope gives a field pointing leftwards.

Summary Box

Key Takeaway: The electric field is the negative spatial rate of change of electric potential. Think of it as the “push” a charge feels due to the “hill” of potential.