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Edexcel A Level (IAL) Physics-4.17 Electric Field between Parallel Plates- Study Notes- New Syllabus

Edexcel A Level (IAL) Physics -4.17 Electric Field between Parallel Plates- Study Notes- New syllabus

Edexcel A Level (IAL) Physics -4.17 Electric Field between Parallel Plates- Study Notes -Edexcel A level Physics – per latest Syllabus.

Key Concepts:

  • be able to use the equation \(E=\dfrac{V}{d}\) for an electric field between parallel plates

Edexcel A level Physics-Study Notes- All Topics

Electric Field Between Parallel Plates \( E = \dfrac{V}{d} \)

When two large, flat, parallel conducting plates are connected to a potential difference, a uniform electric field is produced between them (ignoring edge effects).

Equation for Electric Field Between Parallel Plates

The magnitude of the electric field between the plates is given by:

\( E = \dfrac{V}{d} \)

  • \( E \) = electric field strength (V m⁻¹ or N C⁻¹)
  • \( V \) = potential difference between the plates (V)
  • \( d \) = separation between the plates (m)

Conditions for Using This Equation

  • The plates are large compared to their separation.
  • The plates are parallel.
  • The electric field between the plates is uniform.
  • Edge effects are neglected.

Direction of the Electric Field

  • The electric field points from the positive plate to the negative plate.
  • The field direction is the direction of force on a positive test charge.

Link with Potential Gradient

Electric field strength is the potential gradient:

\( E = -\dfrac{\Delta V}{\Delta x} \)

For a uniform field, the potential changes linearly with distance, giving:

\( E = \dfrac{V}{d} \)

Relationship with Force on a Charge

A charge \( Q \) placed between the plates experiences a force:

\( F = EQ \)

This force is constant anywhere between the plates because the field is uniform.

 Applications

  • Motion of charged particles between plates
  • Electron deflection in cathode-ray tubes
  • Capacitors
  • Particle accelerators

Example (Easy)

Two parallel plates have a potential difference of \( 200\ \mathrm{V} \) and are separated by \( 0.020\ \mathrm{m} \). Calculate the electric field strength between the plates.

▶️ Answer / Explanation

\( E = \dfrac{V}{d} = \dfrac{200}{0.020} = 1.0\times10^{4}\ \mathrm{V\,m^{-1}} \)

Example (Medium)

An electric field of strength \( 3.0\times10^{4}\ \mathrm{V\,m^{-1}} \) exists between two plates. If the plates are \( 0.015\ \mathrm{m} \) apart, find the potential difference.

▶️ Answer / Explanation

\( V = Ed = (3.0\times10^{4})(0.015) = 450\ \mathrm{V} \)

Example (Hard)

Explain what happens to the electric field strength if the potential difference is doubled while the plate separation remains constant.

▶️ Answer / Explanation
  • Electric field strength is proportional to potential difference.
  • Doubling \( V \) doubles \( E \).
  • The field remains uniform.
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