Capacitance of a parallel plate capacitor

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
$L = 1.5 m, A = 1.5e-5 m^{2}$
$\varepsilon_r = 7.5, A_p = 0.0190 m^{2}, d = 0.00080 m$
$V = 130 V$

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
$R = \rho L / A$
Substituting
$R = 0.00e+0(1.5)/1.5e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(7.5)(0.0190)/0.00080 = 1.577e-9 F$
Formula
$Q = C V$
Substituting
$Q = 1.577e-9(130) = 2.050e-7 C$

Answer:

R = 0.000 \Omega, 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
$L = 4.0 m, A = 1.7e-5 m^{2}$
$\varepsilon_r = 6.5, A_p = 0.0180 m^{2}, d = 0.00010 m$
$V = 135 V$

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
$R = \rho L / A$
Substituting
$R = 0.00e+0(4.0)/1.7e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(6.5)(0.0180)/0.00010 = 1.036e-8 F$
Formula
$Q = C V$
Substituting
$Q = 1.036e-8(135) = 1.398e-6 C$

Answer:

R = 0.000 \Omega, 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
$L = 3.5 m, A = 1.0e-6 m^{2}$
$\varepsilon_r = 3.5, A_p = 0.0090 m^{2}, d = 0.00100 m$
$V = 183 V$

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
$R = \rho L / A$
Substituting
$R = 1.00e-6(3.5)/1.0e-6 = 3.5000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(3.5)(0.0090)/0.00100 = 2.789e-10 F$
Formula
$Q = C V$
Substituting
$Q = 2.789e-10(183) = 5.104e-8 C$

Answer:

R = 3.500 \Omega, 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
$L = 3.0 m, A = 1.6e-5 m^{2}$
$\varepsilon_r = 5.5, A_p = 0.0100 m^{2}, d = 0.00080 m$
$V = 69 V$

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
$R = \rho L / A$
Substituting
$R = 0.00e+0(3.0)/1.6e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(5.5)(0.0100)/0.00080 = 6.087e-10 F$
Formula
$Q = C V$
Substituting
$Q = 6.087e-10(69) = 4.200e-8 C$

Answer:

R = 0.000 \Omega, 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
$L = 3.0 m, A = 3.0e-6 m^{2}$
$\varepsilon_r = 5.0, A_p = 0.0160 m^{2}, d = 0.00080 m$
$V = 31 V$

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
$R = \rho L / A$
Substituting
$R = 1.00e-6(3.0)/3.0e-6 = 1.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(5.0)(0.0160)/0.00080 = 8.854e-10 F$
Formula
$Q = C V$
Substituting
$Q = 8.854e-10(31) = 2.745e-8 C$

Answer:

R = 1.000 \Omega, 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
$L = 1.0 m, A = 8.0e-6 m^{2}$
$\varepsilon_r = 5.0, A_p = 0.0140 m^{2}, d = 0.00070 m$
$V = 28 V$

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
$R = \rho L / A$
Substituting
$R = 0.00e+0(1.0)/8.0e-6 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(5.0)(0.0140)/0.00070 = 8.854e-10 F$
Formula
$Q = C V$
Substituting
$Q = 8.854e-10(28) = 2.479e-8 C$

Answer:

R = 0.000 \Omega, 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
$L = 1.5 m, A = 1.1e-5 m^{2}$
$\varepsilon_r = 3.5, A_p = 0.0190 m^{2}, d = 0.00060 m$
$V = 148 V$

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
$R = \rho L / A$
Substituting
$R = 1.00e-6(1.5)/1.1e-5 = 0.1364 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(3.5)(0.0190)/0.00060 = 9.813e-10 F$
Formula
$Q = C V$
Substituting
$Q = 9.813e-10(148) = 1.452e-7 C$

Answer:

R = 0.136 \Omega, 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
$L = 2.5 m, A = 6.0e-6 m^{2}$
$\varepsilon_r = 3.5, A_p = 0.0090 m^{2}, d = 0.00050 m$
$V = 110 V$

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
$R = \rho L / A$
Substituting
$R = 0.00e+0(2.5)/6.0e-6 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(3.5)(0.0090)/0.00050 = 5.578e-10 F$
Formula
$Q = C V$
Substituting
$Q = 5.578e-10(110) = 6.136e-8 C$

Answer:

R = 0.000 \Omega, 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
$L = 2.5 m, A = 8.0e-6 m^{2}$
$\varepsilon_r = 7.5, A_p = 0.0130 m^{2}, d = 0.00100 m$
$V = 152 V$

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
$R = \rho L / A$
Substituting
$R = 1.00e-6(2.5)/8.0e-6 = 0.3125 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(7.5)(0.0130)/0.00100 = 8.633e-10 F$
Formula
$Q = C V$
Substituting
$Q = 8.633e-10(152) = 1.312e-7 C$

Answer:

R = 0.313 \Omega, 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
$L = 3.5 m, A = 1.9e-5 m^{2}$
$\varepsilon_r = 3.5, A_p = 0.0180 m^{2}, d = 0.00050 m$
$V = 151 V$

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
$R = \rho L / A$
Substituting
$R = 1.00e-6(3.5)/1.9e-5 = 0.1842 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(3.5)(0.0180)/0.00050 = 1.116e-9 F$
Formula
$Q = C V$
Substituting
$Q = 1.116e-9(151) = 1.685e-7 C$

Answer:

R = 0.184 \Omega, 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

Capacitance of a parallel plate capacitor

Handbook notes for this section

Core formulas for this FE topic

Worked exam-style examples

Capacitance of a parallel plate capacitor

Handbook notes for this section

Core formulas for this FE topic

C=dC= dC=d

C=capacitanceC = capacitanceC=capacitance

ε=permittivityofmaterial\varepsilon = permittivity of materialε=permittivityofmaterial

A=cross−sectionalareaoftheplatesA = cross-sectional area of the platesA=cross−sectionalareaoftheplates

d=distancebetweentheplatesd = distance between the platesd=distancebetweentheplates

Worked exam-style examples