10.5 Potential Gradient

The potential gradient is a physical quantity that describes how electric potential changes with respect to distance in an electric field. It is a vector quantity that points in the direction of the steepest increase in electric potential, and its magnitude equals the rate of change of potential with distance. This concept is essential for understanding the relationship between electric field and electric potential.

1. Definition and Relationship to Electric Field

The electric field ($\vec{E}$) is directly related to the potential gradient. Specifically, the electric field is the negative of the potential gradient.

Formula: $E=-\frac{\Delta V}{\Delta r}$

• $E$ is the component of the electric field along the direction of displacement $\Delta r$.
• $\Delta V$ is the change in electric potential.
• $\Delta r$ is the small displacement.

Significance of the Negative Sign:

The negative sign indicates that the electric field vector ($\vec{E}$) always points in the direction of the greatest decrease in electric potential. A positive charge placed in an electric field will naturally move from a region of high potential to a region of low potential.

2. Derivation of the Potential Gradient

The relationship can be derived by considering the work done to move a test charge in an electric field.

1. Consider a positive test charge $q_0$ being moved a small distance $\Delta r$ from point A to point B against a uniform electric field $\vec{E}$.
2. The external force (${\vec{F}}_{ext}$) required to move the charge must be equal and opposite to the electric force (${\vec{F}}_E=q_0\vec{E}$), so ${\vec{F}}_{ext}=-q_0\vec{E}$.
3. The work done ($W$) by this external force is: $W=F_{ext}\Delta r=(-q_{0}E)\Delta r$
4. By the definition of electric potential difference, the work done to move a charge is also equal to the charge multiplied by the potential difference: $W=q_0\Delta V$
5. Equating the two expressions for work done: $q_0\Delta V=-q_{0}E\Delta r$
6. Canceling the test charge $q_0$ and rearranging gives the final relationship: $E=-\frac{\Delta V}{\Delta r}$

3. Units of Electric Field Intensity

The potential gradient formula $E=-\Delta V/\Delta r$ shows that the SI unit of electric field intensity can be expressed as volt per metre (V m⁻¹).

This is equivalent to newton per coulomb (N C⁻¹):

$\frac{\text{V}}{\text{m}}=\frac{{\text{J C}}^{-1}}{\text{m}}=\frac{{\text{N m C}}^{-1}}{\text{m}}={\text{N C}}^{-1}$

Since $1\text{ V}=1{\text{ J C}}^{-1}$ and $1\text{ J}=1\text{ N m}$.

4. Analogy with a Hill

A useful analogy is to think of electric potential as the height on a hill (gravitational potential).

• The electric field is like the slope of the hill. It points "downhill," the direction a ball would roll.
• The potential gradient is the steepness of the slope. A large potential gradient (steep slope) corresponds to a strong electric field.
• Moving "uphill" requires work and increases potential. Moving "downhill" releases potential energy.

5. Motion of a Charged Particle in a Uniform Electric Field

A uniform electric field exerts a constant force on a charged particle, causing it to accelerate:

• A positive charge ($+q$) placed in an electric field will experience a force in the same direction as the field ($\vec{F}=q\vec{E}$) and will accelerate from a region of high potential to low potential.
• A negative charge ($-q$) will experience a force in the opposite direction to the field and will accelerate from a region of low potential to high potential.

This is directly analogous to a mass falling under gravity — the charge moves to minimise its potential energy.

6. Example Calculation

Problem: The electric potential at a point is $V=13$ J/C (or 13 Volts), and the magnitude of the electric field at that point is $E=26$ N/C. Assuming the field is created by a single point charge, what is the distance ($r$) from the charge?

Solution:

For a point charge, the potential is $V=kQ/r$ and the electric field is $E=kQ/r^2$. Dividing gives $E=V/r$.

1. Formula: Rearrange to solve for distance: $r=\frac{V}{E}$
2. Substitute values: $r=\frac{13\mathrm{V}}{26\mathrm{V}{\mathrm{m}}^{-1}}=0.5\mathrm{m}$

The distance from the charge is 0.5 m.

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