Charge held by a capacitor

Example 1

Resistivity of a conductor and capacitance of a parallel plate — Charge held by a capacitor

A conductor of resistivity 1.00e-6 Ω·m is 2.0 m long with a cross-section of 1.4e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 4.5), plate area 0.0110 m² and spacing 0.00020 m is charged to 127 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 2.0 m, A = 1.4e-5 m^{2}$
$\varepsilon_r = 4.5, A_p = 0.0110 m^{2}, d = 0.00020 m$
$V = 127 V$

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
$R = 1.00e-6(2.0)/1.4e-5 = 0.1429 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(4.5)(0.0110)/0.00020 = 2.191e-9 F$
Formula
$Q = C V$
Substituting
$Q = 2.191e-9(127) = 2.783e-7 C$

Answer:

R = 0.143 \Omega, C = 2.19e-9 F, Q = 2.78e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 4.87e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Charge held by a capacitor

Example 2

Resistivity of a conductor and capacitance of a parallel plate — Charge held by a capacitor (2)

A conductor of resistivity 0.00e+0 Ω·m is 3.5 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 = 7.0), plate area 0.0030 m² and spacing 0.00060 m is charged to 174 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 3.5 m, A = 1.5e-5 m^{2}$
$\varepsilon_r = 7.0, A_p = 0.0030 m^{2}, d = 0.00060 m$
$V = 174 V$

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
$R = 0.00e+0(3.5)/1.5e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(7.0)(0.0030)/0.00060 = 3.099e-10 F$
Formula
$Q = C V$
Substituting
$Q = 3.099e-10(174) = 5.392e-8 C$

Answer:

R = 0.000 \Omega, C = 3.10e-10 F, Q = 5.39e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 4.43e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Charge held by a capacitor

Example 3

Resistivity of a conductor and capacitance of a parallel plate — Charge held by a capacitor (3)

A conductor of resistivity 0.00e+0 Ω·m is 1.0 m long with a cross-section of 7.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 3.0), plate area 0.0070 m² and spacing 0.00010 m is charged to 73 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 1.0 m, A = 7.0e-6 m^{2}$
$\varepsilon_r = 3.0, A_p = 0.0070 m^{2}, d = 0.00010 m$
$V = 73 V$

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
$R = 0.00e+0(1.0)/7.0e-6 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(3.0)(0.0070)/0.00010 = 1.859e-9 F$
Formula
$Q = C V$
Substituting
$Q = 1.859e-9(73) = 1.357e-7 C$

Answer:

R = 0.000 \Omega, C = 1.86e-9 F, Q = 1.36e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 6.20e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Charge held by a capacitor

Example 4

Resistivity of a conductor and capacitance of a parallel plate — Charge held by a capacitor (4)

A conductor of resistivity 1.00e-6 Ω·m is 2.5 m long with a cross-section of 4.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 4.5), plate area 0.0080 m² and spacing 0.00090 m is charged to 45 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 2.5 m, A = 4.0e-6 m^{2}$
$\varepsilon_r = 4.5, A_p = 0.0080 m^{2}, d = 0.00090 m$
$V = 45 V$

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
$R = 1.00e-6(2.5)/4.0e-6 = 0.6250 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(4.5)(0.0080)/0.00090 = 3.542e-10 F$
Formula
$Q = C V$
Substituting
$Q = 3.542e-10(45) = 1.594e-8 C$

Answer:

R = 0.625 \Omega, C = 3.54e-10 F, Q = 1.59e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 7.87e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Charge held by a capacitor

Example 5

Resistivity of a conductor and capacitance of a parallel plate — Charge held by a capacitor (5)

A conductor of resistivity 1.00e-6 Ω·m is 2.5 m long with a cross-section of 1.0e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 4.0), plate area 0.0020 m² and spacing 0.00020 m is charged to 115 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 2.5 m, A = 1.0e-5 m^{2}$
$\varepsilon_r = 4.0, A_p = 0.0020 m^{2}, d = 0.00020 m$
$V = 115 V$

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
$R = 1.00e-6(2.5)/1.0e-5 = 0.2500 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(4.0)(0.0020)/0.00020 = 3.542e-10 F$
Formula
$Q = C V$
Substituting
$Q = 3.542e-10(115) = 4.073e-8 C$

Answer:

R = 0.250 \Omega, C = 3.54e-10 F, Q = 4.07e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 8.85e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Charge held by a capacitor

Example 6

Resistivity of a conductor and capacitance of a parallel plate — Charge held by a capacitor (6)

A conductor of resistivity 0.00e+0 Ω·m is 1.5 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 = 3.0), plate area 0.0090 m² and spacing 0.00100 m is charged to 15 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 1.5 m, A = 6.0e-6 m^{2}$
$\varepsilon_r = 3.0, A_p = 0.0090 m^{2}, d = 0.00100 m$
$V = 15 V$

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
$R = 0.00e+0(1.5)/6.0e-6 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(3.0)(0.0090)/0.00100 = 2.391e-10 F$
Formula
$Q = C V$
Substituting
$Q = 2.391e-10(15) = 3.586e-9 C$

Answer:

R = 0.000 \Omega, C = 2.39e-10 F, Q = 3.59e-9 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 → Charge held by a capacitor

Example 7

Resistivity of a conductor and capacitance of a parallel plate — Charge held by a capacitor (7)

A conductor of resistivity 0.00e+0 Ω·m is 2.0 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 = 6.0), plate area 0.0180 m² and spacing 0.00090 m is charged to 93 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 2.0 m, A = 8.0e-6 m^{2}$
$\varepsilon_r = 6.0, A_p = 0.0180 m^{2}, d = 0.00090 m$
$V = 93 V$

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
$R = 0.00e+0(2.0)/8.0e-6 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(6.0)(0.0180)/0.00090 = 1.062e-9 F$
Formula
$Q = C V$
Substituting
$Q = 1.062e-9(93) = 9.881e-8 C$

Answer:

R = 0.000 \Omega, C = 1.06e-9 F, Q = 9.88e-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 → Charge held by a capacitor

Example 8

Resistivity of a conductor and capacitance of a parallel plate — Charge held by a capacitor (8)

A conductor of resistivity 0.00e+0 Ω·m is 3.5 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 = 6.0), plate area 0.0190 m² and spacing 0.00040 m is charged to 21 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 3.5 m, A = 1.5e-5 m^{2}$
$\varepsilon_r = 6.0, A_p = 0.0190 m^{2}, d = 0.00040 m$
$V = 21 V$

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
$R = 0.00e+0(3.5)/1.5e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(6.0)(0.0190)/0.00040 = 2.523e-9 F$
Formula
$Q = C V$
Substituting
$Q = 2.523e-9(21) = 5.299e-8 C$

Answer:

R = 0.000 \Omega, C = 2.52e-9 F, Q = 5.30e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 4.21e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Charge held by a capacitor

Example 9

Resistivity of a conductor and capacitance of a parallel plate — Charge held by a capacitor (9)

A conductor of resistivity 1.00e-6 Ω·m is 4.5 m long with a cross-section of 5.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 4.5), plate area 0.0060 m² and spacing 0.00070 m is charged to 20 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 4.5 m, A = 5.0e-6 m^{2}$
$\varepsilon_r = 4.5, A_p = 0.0060 m^{2}, d = 0.00070 m$
$V = 20 V$

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
$R = 1.00e-6(4.5)/5.0e-6 = 0.9000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(4.5)(0.0060)/0.00070 = 3.415e-10 F$
Formula
$Q = C V$
Substituting
$Q = 3.415e-10(20) = 6.830e-9 C$

Answer:

R = 0.900 \Omega, C = 3.42e-10 F, Q = 6.83e-9 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 7.59e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Charge held by a capacitor

Example 10

Resistivity of a conductor and capacitance of a parallel plate — Charge held by a capacitor (10)

A conductor of resistivity 1.00e-6 Ω·m is 3.0 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 = 4.5), plate area 0.0120 m² and spacing 0.00070 m is charged to 33 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 3.0 m, A = 6.0e-6 m^{2}$
$\varepsilon_r = 4.5, A_p = 0.0120 m^{2}, d = 0.00070 m$
$V = 33 V$

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
$R = 1.00e-6(3.0)/6.0e-6 = 0.5000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(4.5)(0.0120)/0.00070 = 6.830e-10 F$
Formula
$Q = C V$
Substituting
$Q = 6.830e-10(33) = 2.254e-8 C$

Answer:

R = 0.500 \Omega, C = 6.83e-10 F, Q = 2.25e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.52e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Charge held by a capacitor

Core formulas for this FE topic

Worked exam-style examples