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

Learning Objective

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

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

Key Concepts

• Electric Potential (V): The work done per unit charge in bringing a test charge from infinity to a point in an electric field, measured in volts (V).
• Electric Field (E): A vector field representing the force per unit positive charge, measured in newtons per coulomb (N C⁻¹).
• Potential Gradient: The spatial rate of change of the potential; mathematically the gradient operator applied to V.

Derivation of the Relationship

Consider a small displacement \$d\mathbf{s}\$ in an electric field. The infinitesimal work \$dW\$ done on a charge \$q\$ is:

\$dW = q\,\mathbf{E}\cdot d\mathbf{s}\$

The corresponding change in potential energy \$dU\$ is related to the change in potential \$dV\$ by \$dU = q\,dV\$. Since \$dU = -dW\$ (work done by the field reduces the potential energy), we have:

\$q\,dV = -q\,\mathbf{E}\cdot d\mathbf{s} \quad\Rightarrow\quad dV = -\mathbf{E}\cdot d\mathbf{s}\$

Dividing by the infinitesimal displacement and recognizing the definition of the gradient gives the vector form:

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

Practical Use

1. Identify the potential function \$V(x,y,z)\$ for the situation.
2. Compute its gradient: \$\nabla V = \left(\frac{\partial V}{\partial x},\frac{\partial V}{\partial y},\frac{\partial V}{\partial z}\right)\$.
3. Apply the negative sign to obtain the electric field: \$\mathbf{E} = -\nabla V\$.
4. Check units: \$V\$ in volts, \$\nabla V\$ in V m⁻¹, giving \$\mathbf{E}\$ in V m⁻¹ = N C⁻¹.

Common Potential Functions

Worked Example

Problem: A point charge \$Q = +5\,\mu\text{C}\$ is placed at the origin. Find the electric field at the point \$(0,\,0,\,0.10\ \text{m})\$ using the potential gradient method.

1. Write the potential due to a point charge:

   \$V(r) = \frac{kQ}{r},\qquad k = 8.99\times10^{9}\ \text{N·m}^{2}\text{C}^{-2}\$

2. Express \$r\$ in terms of the coordinates. Here \$r = \sqrt{x^{2}+y^{2}+z^{2}} = z\$ because \$x=y=0\$.
3. Compute the gradient:

   \$\$\nabla V = \frac{dV}{dr}\,\hat{r}

   = -\frac{kQ}{r^{2}}\,\hat{r}\$\$

4. Apply \$\mathbf{E} = -\nabla V\$:

   \$\mathbf{E} = \frac{kQ}{r^{2}}\,\hat{r}\$

5. Insert the numbers:

   \$\$r = 0.10\ \text{m},\quad

   \mathbf{E} = \frac{8.99\times10^{9}\times5\times10^{-6}}{(0.10)^{2}}\hat{z}

   = 4.5\times10^{6}\ \text{N·C}^{-1}\,\hat{z}\$\$

Common Mistakes to Avoid

• Forgetting the negative sign: \$\mathbf{E}\$ points in the direction of decreasing potential.
• Mixing up scalar and vector quantities; the gradient produces a vector.
• Using inconsistent reference points for potential (always state the zero‑potential reference).
• Neglecting unit conversion, especially when \$k\$ is expressed in SI units.

Suggested Diagram

Summary

The electric field at any point can be obtained directly from the spatial variation of the electric potential using the simple vector relation \$\mathbf{E} = -\nabla V\$. Mastery of this relationship enables quick conversion between scalar potential problems and vector field problems, a skill essential for A‑Level examinations.