Capacitance of a parallel plate capacitor

Materials Science · FE Reference Handbook section

Materials Science

6 formulas

10 exam-style examples

~57 min

All Materials Science lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Capacitance of a parallel plate capacitor within Materials Science. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what capacitance of a parallel plate capacitor describes physically and when it applies.
State every one of the 6 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: percent versus fraction in composition and strain.

Lecture

Why this section exists. Capacitance of a parallel plate capacitor is the part of Materials Science that lets you connect a steel, concrete or polymer specimen under test to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a definition, a phase-diagram read, or a one-line property calculation. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. percent versus fraction in composition and strain. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Concrete cylinder under axial load in a compression testing machine. — Photo 1. Where this shows up in practice: capacitance of a parallel plate capacitor.
Wikimedia Commons, public domain

Materials Science — Capacitance of a parallel plate capacitor: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a steel, concrete or polymer specimen under test. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 6 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

C	Quantity produced by "C= d" — read its definition and unit from the handbook line directly above the equation.
ε	Quantity produced by "ε = permittivity of material" — read its definition and unit from the handbook line directly above the equation.
A	Quantity produced by "A = cross-sectional area of the plates" — read its definition and unit from the handbook line directly above the equation.
d	Quantity produced by "d = distance between the plates" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

where
Resistivity: The material property that determines the resistance of a resistor

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Resistivity of a conductor and capacitance of a parallel plate — Capacitance of a parallel plate capacitor

A conductor of resistivity 0.00e+0 Ω·m is 1.5 m long with a cross-section of 1.5e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 7.5), plate area 0.0190 m² and spacing 0.00080 m is charged to 130 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
L = 1.5 m, A = 1.5e-5 m²
ε_r = 7.5, A_p = 0.0190 m², d = 0.00080 m
V = 130 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 0.00 e + 0 (1.5) /1.5 e - 5 = 0.0000Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (7.5) (0.0190) /0.00080 = 1.577 e - 9 F$
Formula
$Q = C V$
Substituting
$Q = 1.577 e - 9 (130) = 2.050 e - 7 C$

Answer: R = 0.000 Ω, C = 1.58e-9 F, Q = 2.05e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 2.10e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Capacitance of a parallel plate capacitor

Example 2

Resistivity of a conductor and capacitance of a parallel plate — Capacitance of a parallel plate capacitor (2)

A conductor of resistivity 0.00e+0 Ω·m is 4.0 m long with a cross-section of 1.7e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 6.5), plate area 0.0180 m² and spacing 0.00010 m is charged to 135 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
L = 4.0 m, A = 1.7e-5 m²
ε_r = 6.5, A_p = 0.0180 m², d = 0.00010 m
V = 135 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 0.00 e + 0 (4.0) /1.7 e - 5 = 0.0000Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (6.5) (0.0180) /0.00010 = 1.036 e - 8 F$
Formula
$Q = C V$
Substituting
$Q = 1.036 e - 8 (135) = 1.398 e - 6 C$

Answer: R = 0.000 Ω, C = 1.04e-8 F, Q = 1.40e-6 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.59e-9 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Capacitance of a parallel plate capacitor

Example 3

Resistivity of a conductor and capacitance of a parallel plate — Capacitance of a parallel plate capacitor (3)

A conductor of resistivity 1.00e-6 Ω·m is 3.5 m long with a cross-section of 1.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 3.5), plate area 0.0090 m² and spacing 0.00100 m is charged to 183 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
L = 3.5 m, A = 1.0e-6 m²
ε_r = 3.5, A_p = 0.0090 m², d = 0.00100 m
V = 183 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 1.00 e - 6 (3.5) /1.0 e - 6 = 3.5000Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (3.5) (0.0090) /0.00100 = 2.789 e - 10 F$
Formula
$Q = C V$
Substituting
$Q = 2.789 e - 10 (183) = 5.104 e - 8 C$

Answer: R = 3.500 Ω, C = 2.79e-10 F, Q = 5.10e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 7.97e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Capacitance of a parallel plate capacitor

Example 4

Resistivity of a conductor and capacitance of a parallel plate — Capacitance of a parallel plate capacitor (4)

A conductor of resistivity 0.00e+0 Ω·m is 3.0 m long with a cross-section of 1.6e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 5.5), plate area 0.0100 m² and spacing 0.00080 m is charged to 69 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
L = 3.0 m, A = 1.6e-5 m²
ε_r = 5.5, A_p = 0.0100 m², d = 0.00080 m
V = 69 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 0.00 e + 0 (3.0) /1.6 e - 5 = 0.0000Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (5.5) (0.0100) /0.00080 = 6.087 e - 10 F$
Formula
$Q = C V$
Substituting
$Q = 6.087 e - 10 (69) = 4.200 e - 8 C$

Answer: R = 0.000 Ω, C = 6.09e-10 F, Q = 4.20e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.11e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Capacitance of a parallel plate capacitor

Example 5

Resistivity of a conductor and capacitance of a parallel plate — Capacitance of a parallel plate capacitor (5)

A conductor of resistivity 1.00e-6 Ω·m is 3.0 m long with a cross-section of 3.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 5.0), plate area 0.0160 m² and spacing 0.00080 m is charged to 31 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
L = 3.0 m, A = 3.0e-6 m²
ε_r = 5.0, A_p = 0.0160 m², d = 0.00080 m
V = 31 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 1.00 e - 6 (3.0) /3.0 e - 6 = 1.0000Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (5.0) (0.0160) /0.00080 = 8.854 e - 10 F$
Formula
$Q = C V$
Substituting
$Q = 8.854 e - 10 (31) = 2.745 e - 8 C$

Answer: R = 1.000 Ω, C = 8.85e-10 F, Q = 2.74e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.77e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Capacitance of a parallel plate capacitor

Example 6

Resistivity of a conductor and capacitance of a parallel plate — Capacitance of a parallel plate capacitor (6)

A conductor of resistivity 0.00e+0 Ω·m is 1.0 m long with a cross-section of 8.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 5.0), plate area 0.0140 m² and spacing 0.00070 m is charged to 28 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
L = 1.0 m, A = 8.0e-6 m²
ε_r = 5.0, A_p = 0.0140 m², d = 0.00070 m
V = 28 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 0.00 e + 0 (1.0) /8.0 e - 6 = 0.0000Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (5.0) (0.0140) /0.00070 = 8.854 e - 10 F$
Formula
$Q = C V$
Substituting
$Q = 8.854 e - 10 (28) = 2.479 e - 8 C$

Answer: R = 0.000 Ω, C = 8.85e-10 F, Q = 2.48e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.77e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Capacitance of a parallel plate capacitor

Example 7

Resistivity of a conductor and capacitance of a parallel plate — Capacitance of a parallel plate capacitor (7)

A conductor of resistivity 1.00e-6 Ω·m is 1.5 m long with a cross-section of 1.1e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 3.5), plate area 0.0190 m² and spacing 0.00060 m is charged to 148 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
L = 1.5 m, A = 1.1e-5 m²
ε_r = 3.5, A_p = 0.0190 m², d = 0.00060 m
V = 148 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 1.00 e - 6 (1.5) /1.1 e - 5 = 0.1364Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (3.5) (0.0190) /0.00060 = 9.813 e - 10 F$
Formula
$Q = C V$
Substituting
$Q = 9.813 e - 10 (148) = 1.452 e - 7 C$

Answer: R = 0.136 Ω, C = 9.81e-10 F, Q = 1.45e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 2.80e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Capacitance of a parallel plate capacitor

Example 8

Resistivity of a conductor and capacitance of a parallel plate — Capacitance of a parallel plate capacitor (8)

A conductor of resistivity 0.00e+0 Ω·m is 2.5 m long with a cross-section of 6.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 3.5), plate area 0.0090 m² and spacing 0.00050 m is charged to 110 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
L = 2.5 m, A = 6.0e-6 m²
ε_r = 3.5, A_p = 0.0090 m², d = 0.00050 m
V = 110 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 0.00 e + 0 (2.5) /6.0 e - 6 = 0.0000Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (3.5) (0.0090) /0.00050 = 5.578 e - 10 F$
Formula
$Q = C V$
Substituting
$Q = 5.578 e - 10 (110) = 6.136 e - 8 C$

Answer: R = 0.000 Ω, C = 5.58e-10 F, Q = 6.14e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.59e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Capacitance of a parallel plate capacitor

Example 9

Resistivity of a conductor and capacitance of a parallel plate — Capacitance of a parallel plate capacitor (9)

A conductor of resistivity 1.00e-6 Ω·m is 2.5 m long with a cross-section of 8.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 7.5), plate area 0.0130 m² and spacing 0.00100 m is charged to 152 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
L = 2.5 m, A = 8.0e-6 m²
ε_r = 7.5, A_p = 0.0130 m², d = 0.00100 m
V = 152 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 1.00 e - 6 (2.5) /8.0 e - 6 = 0.3125Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (7.5) (0.0130) /0.00100 = 8.633 e - 10 F$
Formula
$Q = C V$
Substituting
$Q = 8.633 e - 10 (152) = 1.312 e - 7 C$

Answer: R = 0.313 Ω, C = 8.63e-10 F, Q = 1.31e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.15e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Capacitance of a parallel plate capacitor

Example 10

Resistivity of a conductor and capacitance of a parallel plate — Capacitance of a parallel plate capacitor (10)

A conductor of resistivity 1.00e-6 Ω·m is 3.5 m long with a cross-section of 1.9e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 3.5), plate area 0.0180 m² and spacing 0.00050 m is charged to 151 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
L = 3.5 m, A = 1.9e-5 m²
ε_r = 3.5, A_p = 0.0180 m², d = 0.00050 m
V = 151 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 1.00 e - 6 (3.5) /1.9 e - 5 = 0.1842Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (3.5) (0.0180) /0.00050 = 1.116 e - 9 F$
Formula
$Q = C V$
Substituting
$Q = 1.116 e - 9 (151) = 1.685 e - 7 C$

Answer: R = 0.184 Ω, C = 1.12e-9 F, Q = 1.68e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 3.19e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Capacitance of a parallel plate capacitor

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a steel, concrete or polymer specimen under test, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Capacitance of a parallel plate capacitor contains 6 relations; you must be able to find this page in under 15 seconds.
Exam style: a definition, a phase-diagram read, or a one-line property calculation.
Unit rule: percent versus fraction in composition and strain.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

percent versus fraction in composition and strain
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.