recall and use E = ∆V / ∆d to calculate the field strength of the uniform field between charged parallel plates

Uniform Electric Fields

Learning Objective

Recall and use the relationship

\$E = \frac{\Delta V}{\Delta d}\$

to calculate the electric field strength of a uniform field between charged parallel plates.

Key Concepts

• Electric field (E): a vector quantity representing the force per unit charge, measured in newtons per coulomb (N C⁻¹) or volts per metre (V m⁻¹).
• Potential difference (ΔV): the work done per unit charge in moving between two points, measured in volts (V).
• Separation distance (Δd): the straight‑line distance between the two points where the potential is measured, measured in metres (m).
• Uniform field: the field has the same magnitude and direction at every point between the plates.

Derivation of \$E = \dfrac{\Delta V}{\Delta d}\$ for Parallel Plates

Consider two large, parallel conducting plates separated by a distance \$d\$ and carrying equal and opposite charges. The electric field between them is uniform and directed from the positive to the negative plate.

The work \$W\$ done in moving a test charge \$q\$ across the field from the positive to the negative plate is:

\$W = q\,\Delta V\$

But work is also the product of force and displacement:

\$W = F\,\Delta d = qE\,\Delta d\$

Equating the two expressions for \$W\$ and cancelling \$q\$ gives:

\$E = \frac{\Delta V}{\Delta d}\$

Since the field is uniform, \$E\$ is constant throughout the region.

Applying the Formula

When the field is uniform, the magnitude of the electric field can be found directly from the measured potential difference and the plate separation.

1. Measure or be given the potential difference \$\Delta V\$ between the plates (in volts).
2. Measure or be given the separation distance \$\Delta d\$ (in metres).
3. Calculate \$E\$ using \$E = \Delta V / \Delta d\$ (units V m⁻¹, which are equivalent to N C⁻¹).

Example Calculation

Two parallel plates are 5.0 cm apart and have a potential difference of 250 V. Find the magnitude of the uniform electric field between them.

Solution:

• Convert distance to metres: \$\Delta d = 5.0\ \text{cm} = 0.050\ \text{m}\$.
• Use the formula:

\$E = \frac{250\ \text{V}}{0.050\ \text{m}} = 5.0\times10^{3}\ \text{V m}^{-1}\$

Thus \$E = 5.0\times10^{3}\ \text{N C}^{-1}\$, directed from the positive to the negative plate.

Common Mistakes to Avoid

• Forgetting to convert the plate separation to metres before using the formula.
• Using the absolute value of the potential difference without considering the direction of the field (the field points from higher to lower potential).
• Assuming the formula applies to non‑uniform fields; it is only valid when \$E\$ is constant between the points considered.
• Mixing up units: \$E\$ expressed in V m⁻¹ is equivalent to N C⁻¹, but the numerical value must be consistent with the chosen units.

Practice Questions

1. A pair of parallel plates are 12 mm apart and have a potential difference of 480 V. Calculate the electric field strength.
2. If the uniform electric field between two plates is \$2.5\times10^{4}\ \text{V m}^{-1}\$ and the separation is 0.020 m, what is the potential difference between the plates?
3. A proton (charge \$+e\$) enters a uniform field of \$1.0\times10^{3}\ \text{N C}^{-1}\$ between parallel plates. If the field extends over a distance of 0.10 m, what is the change in electric potential energy of the proton?

Answers to Practice Questions

Suggested Diagram

Summary

• The magnitude of a uniform electric field between parallel plates is given by \$E = \Delta V / \Delta d\$.
• Ensure distance is in metres and potential difference in volts for a correct numerical result.
• The direction of the field is from higher to lower electric potential.
• This relationship is a cornerstone for many A‑Level problems involving capacitors, particle motion in fields, and energy calculations.