OCR · A-Level · Physics
Course version: H556 · Version from Sept 2015
(b) parallel plate capacitor; permittivity; C = ε₀A/d; C = εA/d; ε = εrε₀
Full specification wording
Uniform electric field: (b) parallel plate capacitor; permittivity; C = ε₀A/d; C = εA/d; ε = εrε₀
A parallel plate capacitor consists of two conducting plates separated by a distance d, with area A overlap.
Checked against the OCR H556 specification (opens in a new tab) on .
- Point 2 of 3
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
A parallel plate capacitor consists of two conducting plates separated by a distance d, with area A overlap. Its capacitance C is given by C = ε₀A/d for a vacuum or air, and C = εA/d when a dielectric of permittivity ε is present. The permittivity ε is related to the permittivity of free space ε₀ by ε = εrε₀, where εr is the relative permittivity (dielectric constant) of the material.
For example: If A = 0.020 m², d = 1.0 mm = 1.0 × 10⁻³ m, and εr = 2.0, then ε = 2.0 × 8.85 × 10⁻¹² = 1.77 × 10⁻¹¹ F m⁻¹, and C = (1.77 × 10⁻¹¹ × 0.020) / (1.0 × 10⁻³) = 3.54 × 10⁻¹⁰ F. You must be able to use these equations and explain how area, separation and dielectric affect capacitance.
What earns the marks
- State the equation for capacitance of a parallel plate capacitor: C = ε₀A/d or C = εA/d.
- Define permittivity and relative permittivity, and use ε = εrε₀.
- Explain that capacitance increases with plate area and permittivity, and decreases with separation.
- Calculate capacitance, area, separation or permittivity using the equations.
- Recognise that ε₀ is the permittivity of free space and εr is dimensionless.
Common mistakes and how to put them right
Using the diameter instead of the area of the plates.
Put it right: Calculate area from the radius if given diameter, and ensure consistent units.
Forgetting to convert separation to metres.
Put it right: Convert mm or cm to m before substitution.
Confusing ε and ε₀.
Put it right: ε is the permittivity of the dielectric, ε₀ is the permittivity of free space, and ε = εrε₀.
Assuming εr has units.
Put it right: Relative permittivity is a dimensionless ratio.
Exam tips
- Write down the equation and rearrange before substituting numbers.
- Check that all quantities are in SI units: area in m², separation in m, permittivity in F m⁻¹.
- Remember that adding a dielectric increases capacitance by a factor of εr.
Try the lesson
The start of the Uniform electric field: (b) parallel plate capacitor; permittivity; C = ε₀A/d; C = εA/d; ε = εrε₀ lesson, made from checked teaching notes for this specification point.
Listen
A parallel plate capacitor consists of two conducting plates separated by a distance d, with area A overlap. Its capacitance C is given by C = ε₀A/d for a vacuum or air, and C = εA/d when a dielectric of permittivity ε is present. The permittivity ε is related to the permittivity of free space ε₀ by ε = εrε₀, where εr is the relative permittivity (dielectric constant) of the material. For example, if A = 0.020 m², d = 1.0 mm = 1.0 × 10⁻³ m, and εr = 2.0, then ε = 2.0 × 8.85 × 10⁻¹² = 1.77 × 10⁻¹¹ F m⁻¹, and C = (1.77 × 10⁻¹¹ × 0.020) / (1.0 × 10⁻³) = 3.54 × 10⁻¹⁰ F. You must be able to use these equations and explain how area, separation and dielectric affect capacitance. A strong answer earns marks for these points. State the equation for capacitance of a parallel plate capacitor: C = ε₀A/d or C = εA/d. Define permittivity and relative permittivity, and use ε = εrε₀. Explain that capacitance increases with plate area and permittivity, and decreases with separation. Calculate capacitance, area, separation or permittivity using the equations. A common slip: Using the diameter instead of the area of the plates. Instead, calculate area from the radius if given diameter, and ensure consistent units. Exam tip: Write down the equation and rearrange before substituting numbers.
See it
Key words
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Parallel plate capacitor
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Permittivity
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Relative permittivity
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Factors affecting capacitance
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See every definition
- Parallel plate capacitor
- Consists of two conducting plates separated by a distance d, with area A overlap.
- Permittivity
- Related to the permittivity of free space ε₀ by ε = εrε₀, where εr is the relative permittivity of the material.
- Relative permittivity
- The dielectric constant of the material, denoted εr, which is dimensionless.
- Factors affecting capacitance
- Capacitance increases with plate area and permittivity, and decreases with separation.
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