A conductor of resistivity 0.00e+0 Ω·m is 1.5 m long with a cross-section of 1.8e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 5.5), plate area 0.0140 m² and spacing 0.00080 m is charged to 78 V — find its capacitance and stored charge.

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

1. Formula

   $$R=\rho L/A$$

2. Substituting

   $$R=0.00e+0(1.5)/1.8e-5=0.0000\Omega$$

3. Formula

   $$C={\epsilon }_0{\epsilon }_{r}A/d$$

4. Substituting

   $$C=(8.854\times 10^{-12})(5.5)(0.0140)/0.00080=8.522e-10F$$

5. Formula

   $$Q=CV$$

6. Substituting

   $$Q=8.522e-10(78)=6.647e-8C$$

Why the other options are there

• R = 0.000000 Ω (L and A swapped)
• C = 1.55e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

A conductor of resistivity 0.00e+0 Ω·m is 2.5 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 = 7.0), plate area 0.0010 m² and spacing 0.00020 m is charged to 62 V — find its capacitance and stored charge.

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

1. Formula

   $$R=\rho L/A$$

2. Substituting

   $$R=0.00e+0(2.5)/1.7e-5=0.0000\Omega$$

3. Formula

   $$C={\epsilon }_0{\epsilon }_{r}A/d$$

4. Substituting

   $$C=(8.854\times 10^{-12})(7.0)(0.0010)/0.00020=3.099e-10F$$

5. Formula

   $$Q=CV$$

6. Substituting

   $$Q=3.099e-10(62)=1.921e-8C$$

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 → Possible Cathode Reactions (Reduction)

A conductor of resistivity 0.00e+0 Ω·m is 0.5 m long with a cross-section of 9.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 6.0), plate area 0.0050 m² and spacing 0.00100 m is charged to 92 V — find its capacitance and stored charge.

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

1. Formula

   $$R=\rho L/A$$

2. Substituting

   $$R=0.00e+0(0.5)/9.0e-6=0.0000\Omega$$

3. Formula

   $$C={\epsilon }_0{\epsilon }_{r}A/d$$

4. Substituting

   $$C=(8.854\times 10^{-12})(6.0)(0.0050)/0.00100=2.656e-10F$$

5. Formula

   $$Q=CV$$

6. Substituting

   $$Q=2.656e-10(92)=2.444e-8C$$

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 → Possible Cathode Reactions (Reduction)

A conductor of resistivity 0.00e+0 Ω·m is 1.0 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 = 5.5), plate area 0.0130 m² and spacing 0.00040 m is charged to 169 V — find its capacitance and stored charge.

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

1. Formula

   $$R=\rho L/A$$

2. Substituting

   $$R=0.00e+0(1.0)/5.0e-6=0.0000\Omega$$

3. Formula

   $$C={\epsilon }_0{\epsilon }_{r}A/d$$

4. Substituting

   $$C=(8.854\times 10^{-12})(5.5)(0.0130)/0.00040=1.583e-9F$$

5. Formula

   $$Q=CV$$

6. Substituting

   $$Q=1.583e-9(169)=2.675e-7C$$

Why the other options are there

• R = 0.000000 Ω (L and A swapped)
• C = 2.88e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

A conductor of resistivity 0.00e+0 Ω·m is 1.5 m long with a cross-section of 2.0e-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 0.00020 m is charged to 102 V — find its capacitance and stored charge.

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

1. Formula

   $$R=\rho L/A$$

2. Substituting

   $$R=0.00e+0(1.5)/2.0e-6=0.0000\Omega$$

3. Formula

   $$C={\epsilon }_0{\epsilon }_{r}A/d$$

4. Substituting

   $$C=(8.854\times 10^{-12})(4.5)(0.0060)/0.00020=1.195e-9F$$

5. Formula

   $$Q=CV$$

6. Substituting

   $$Q=1.195e-9(102)=1.219e-7C$$

Why the other options are there

• R = 0.000000 Ω (L and A swapped)
• C = 2.66e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

A conductor of resistivity 0.00e+0 Ω·m is 1.5 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 = 6.0), plate area 0.0040 m² and spacing 0.00070 m is charged to 8 V — find its capacitance and stored charge.

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

1. Formula

   $$R=\rho L/A$$

2. Substituting

   $$R=0.00e+0(1.5)/1.9e-5=0.0000\Omega$$

3. Formula

   $$C={\epsilon }_0{\epsilon }_{r}A/d$$

4. Substituting

   $$C=(8.854\times 10^{-12})(6.0)(0.0040)/0.00070=3.036e-10F$$

5. Formula

   $$Q=CV$$

6. Substituting

   $$Q=3.036e-10(8)=2.429e-9C$$

Why the other options are there

• R = 0.000000 Ω (L and A swapped)
• C = 5.06e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

A conductor of resistivity 1.00e-6 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 = 2.0), plate area 0.0080 m² and spacing 0.00040 m is charged to 185 V — find its capacitance and stored charge.

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

1. Formula

   $$R=\rho L/A$$

2. Substituting

   $$R=1.00e-6(4.0)/1.7e-5=0.2353\Omega$$

3. Formula

   $$C={\epsilon }_0{\epsilon }_{r}A/d$$

4. Substituting

   $$C=(8.854\times 10^{-12})(2.0)(0.0080)/0.00040=3.542e-10F$$

5. Formula

   $$Q=CV$$

6. Substituting

   $$Q=3.542e-10(185)=6.552e-8C$$

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 → Possible Cathode Reactions (Reduction)

A conductor of resistivity 1.00e-6 Ω·m is 1.0 m long with a cross-section of 2.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 4.0), plate area 0.0100 m² and spacing 0.00010 m is charged to 72 V — find its capacitance and stored charge.

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

1. Formula

   $$R=\rho L/A$$

2. Substituting

   $$R=1.00e-6(1.0)/2.0e-6=0.5000\Omega$$

3. Formula

   $$C={\epsilon }_0{\epsilon }_{r}A/d$$

4. Substituting

   $$C=(8.854\times 10^{-12})(4.0)(0.0100)/0.00010=3.542e-9F$$

5. Formula

   $$Q=CV$$

6. Substituting

   $$Q=3.542e-9(72)=2.550e-7C$$

Why the other options are there

• R = 0.000000 Ω (L and A swapped)
• C = 8.85e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

A conductor of resistivity 1.00e-6 Ω·m is 1.0 m long with a cross-section of 1.8e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 2.5), plate area 0.0040 m² and spacing 0.00050 m is charged to 93 V — find its capacitance and stored charge.

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

1. Formula

   $$R=\rho L/A$$

2. Substituting

   $$R=1.00e-6(1.0)/1.8e-5=0.0556\Omega$$

3. Formula

   $$C={\epsilon }_0{\epsilon }_{r}A/d$$

4. Substituting

   $$C=(8.854\times 10^{-12})(2.5)(0.0040)/0.00050=1.771e-10F$$

5. Formula

   $$Q=CV$$

6. Substituting

   $$Q=1.771e-10(93)=1.647e-8C$$

Why the other options are there

• R = 0.000000 Ω (L and A swapped)
• C = 7.08e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

A conductor of resistivity 0.00e+0 Ω·m is 0.5 m long with a cross-section of 1.2e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 8.0), plate area 0.0030 m² and spacing 0.00020 m is charged to 199 V — find its capacitance and stored charge.

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

1. Formula

   $$R=\rho L/A$$

2. Substituting

   $$R=0.00e+0(0.5)/1.2e-5=0.0000\Omega$$

3. Formula

   $$C={\epsilon }_0{\epsilon }_{r}A/d$$

4. Substituting

   $$C=(8.854\times 10^{-12})(8.0)(0.0030)/0.00020=1.062e-9F$$

5. Formula

   $$Q=CV$$

6. Substituting

   $$Q=1.062e-9(199)=2.114e-7C$$

Why the other options are there

• R = 0.000000 Ω (L and A swapped)
• C = 1.33e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)