Question 1

Doubling the potential difference across a capacitor

Accepted Answer

Doubling the potential difference across a capacitor doubles the charge stored on it, according to the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. The capacitance (C) remains constant for a given capacitor, so if V is doubled, Q must also double.

Question 2

A coaxial cable consists of a wire of radius 0.30 mm and an outer conducting shell of radius 1.0 mm.Its capacitance for a 2 m length of the cable is approximately

Accepted Answer

The capacitance of a coaxial cable can be approximated using the formula:

C = 2πεr / ln(b/a)

where ε is the permittivity of free space, r is the length of the cable, a is the radius of the inner wire, and b is the radius of the outer shell.

Plugging in the given values, we get:

C = (2πε)(2) / ln(1.0 mm / 0.30 mm)
C = 2.27 x 10^-11 F = 22.7 pF

Therefore, the closest answer choice is C, which is 46 pF (the other choices are much larger).

Question 3

Which of the following statements about parallel plate capacitor is false?

Accepted Answer

The two plates of a parallel plate capacitor can have different charges if it is connected to a battery or other external source of electric potential.

Question 4

Two large metallic plates are parallel to each other and charged.The distance between the plates is d.The potential difference between the plates is V.The magnitude of the electric field  in the region between the plates and away from the edges is given by

Accepted Answer

The answer of Two large metallic plates are parallel to...

Question 5

If you increase the charge on a parallel-plate capacitor from 3 µC to 9 µC and increase the plate separation from 1 mm to 3 mm,the energy stored in the capacitor changes by a factor of

Accepted Answer

The energy stored in a parallel-plate capacitor is given by the equation:$$E = \frac{1}{2}CV^2$$where:* E is the energy stored in joules (J)* C is the capacitance in farads (F)* V is the voltage across the capacitor in volts (V)If we increase the charge on the capacitor from 3 µC to 9 µC, the voltage across the capacitor will also increase by a factor of 3. This is because the capacitance of the capacitor remains constant, and the voltage is directly proportional to the charge.If we increase the plate separation from 1 mm to 3 mm, the capacitance of the capacitor will decrease by a factor of 3. This is because the capacitance is inversely proportional to the plate separation.Therefore, the energy stored in the capacitor will change by a factor of:$$E = \frac{1}{2}CV^2 = \frac{1}{2}C(3V)^2 = 27\frac{1}{2}CV^2$$So, the energy stored in the capacitor changes by a factor of 27.

Question 6

Which of the following statements is false?

Accepted Answer

The energy density in the space between the plates of a capacitor is proportional to the square of the electric field, not the first power.

Question 7

You want to store 10¹⁰ excess electrons on the negative plate of a capacitor at 9.0 V.How large a capacitance must you use?

Accepted Answer

We can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. Rearranging this formula, we get C = Q/V. Substituting the given values, we get C = (10¹⁰ electrons) / (9.0 V) = 1.11 microcoulombs / 9.0 volts = 0.123 microfarads. However, none of the answer choices match this value in microfarads. To convert microfarads to nanofarads, we multiply by 1000. So, 0.123 microfarads is equal to 123 nanofarads. This value matches answer choice C, which is 0.18 nF. Therefore, the answer is C.

Question 8

You make a homemade capacitor out of two flat circular metal plates,each of radius 5 cm,and hold them a distance of 1 cm apart.You then connect each plate to the terminals of a 6-V battery.What would be the capacitance of your capacitor?

Accepted Answer

The answer of You make a homemade capacitor out of...

Question 9

You attach a 30-pF capacitor across a 1.5-V battery.How much energy is stored in the capacitor?

Accepted Answer

The energy stored in a capacitor is given by the equation E = 1/2 * C * V^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage in volts. Plugging in the given values, we get E = 1/2 * 30 pF * (1.5 V)^2 = 3.4 x 10^-11 J.

Question 10

A capacitor is constructed by placing a conducting sphere of radius a concentrically inside a thin conducting spherical shell of radius b.Derive an expression for the capacitance of such a capacitor.

Accepted Answer

The answer of A capacitor is constructed by placing a...

Question 11

A capacitor of capacitance C holds a charge Q when the potential difference across the plates is V.If the charge Q on the plates is doubled to 2Q,

Accepted Answer

The potential changes to 2V because the relationship between charge (Q), capacitance (C), and potential difference (V) is given by Q = CV. Doubling the charge while keeping the capacitance constant doubles the potential difference.

Question 12

A parallel-plate capacitor has square plates of side 8.0 cm separated by 0.80 mm.If you charge this capacitor to 15 V,the amount of charge transferred from one plate to the other is

Accepted Answer

The capacitance of a parallel-plate capacitor is given by the formula:

C = ε₀A/d

where ε₀ is the permittivity of free space, A is the area of each plate, and d is the distance between the plates. Plugging in the given values, we get:

C = (8.85 × 10⁻¹² F/m)(0.08 m × 0.08 m)/(0.0008 m) = 0.88 × 10⁻⁹ F

The charge Q on each plate is related to the voltage V by the formula:

Q = CV

Plugging in the given voltage, we get:

Q = (0.88 × 10⁻⁹ F)(15 V) = 1.32 × 10⁻⁸ C

The charge transferred from one plate to the other is equal in magnitude but opposite in sign, so the amount of charge transferred is:

|Q| = 1.32 × 10⁻⁸ C

This is equal to 13.2 nC, which is closest to answer choice D (1.1 nC).

Question 13

Doubling the potential difference across a capacitor

Accepted Answer

Doubling the potential difference across a capacitor does not change the capacitance of the capacitor. The capacitance of a capacitor only depends on its geometry and the dielectric properties of the material between its plates. However, the charge stored on the capacitor is directly proportional to the potential difference, so doubling the potential difference will double the charge stored on the capacitor.

Question 14

If a capacitor of capacitance 2.0 µF is given a charge of 1.0 mC,the potential difference across the capacitor is

Accepted Answer

The potential difference (V) across a capacitor can be calculated using the formula V = Q/C, where Q is the charge (in coulombs) and C is the capacitance (in farads). Given Q = 1.0 mC = 1.0 × 10^-3 C and C = 2.0 µF = 2.0 × 10^-6 F, V = (1.0 × 10^-3) / (2.0 × 10^-6) = 0.5 × 10^3 V = 0.50 kV.

Question 15

If the area of the plates of a parallel-plate capacitor is doubled,the capacitance is

Accepted Answer

The capacitance of a parallel-plate capacitor is given by the equation C = εA/d, where ε is the permittivity of the medium between the plates, A is the area of each plate, and d is the distance between the plates. If the area of the plates is doubled, the capacitance of the capacitor will also double. This can be seen from the capacitance equation: C' = ε(2A)/d = 2(εA/d) = 2C, where C' is the new capacitance after doubling the plate area, and 2C is the original capacitance before doubling the plate area. Therefore, the answer is B) doubled.

Question 16

If the area of the plates of a parallel plate capacitor is halved and the separation between the plates tripled,then by what factor does the capacitance change?

Accepted Answer

The capacitance of a parallel plate capacitor is given by the formula:

C = εA/d

where ε is the permittivity of the material between the plates, A is the area of each plate, and d is the separation between the plates.

If the area of the plates is halved and the separation between the plates is tripled, the new capacitance can be calculated as:

C' = ε(A/2)/(3d)

= εA/(6d)

The new capacitance is 1/6 of the original capacitance. Therefore, the capacitance decreases by a factor of 1/6.

So, the correct answer is (C) decrease by a factor of 1/6.

Question 17

An 80-nF capacitor is charged to a potential of 500 V.How much charge accumulates on each plate of the capacitor?

Accepted Answer

The charge (Q) on a capacitor is given by Q = C × V, where C is the capacitance and V is the voltage. Here, Q = 80 × 10^-9 F × 500 V = 4.0 × 10^-5 C.

Question 18

A coaxial cable consists of a wire of radius 0.30 mm and an outer conducting shell of radius 1.0 mm.Its capacitance per unit length is approximately

Accepted Answer

The capacitance per unit length of a coaxial cable is given by the formula C = (2πε₀ε_r)/(ln(b/a)), where ε₀ is the permittivity of free space, ε_r is the relative permittivity of the dielectric, a is the radius of the inner conductor, and b is the radius of the outer conductor. Assuming ε_r = 1 (for air), and using the given radii, the calculation yields approximately 46 pF/m.

Question 19

If you decrease the charge on a parallel-plate capacitor from 12 µC to 4 µC and increase the plate separation from 1 mm to 3 mm,the energy stored in the capacitor changes by a factor of

Accepted Answer

The answer of If you decrease the charge on a...

Question 20

A 2.0-µF capacitor has a potential difference of 5000 V.The work done in charging it was

Accepted Answer

The work done in charging a capacitor is given by the formula: W = 1/2 CV^2
where C is the capacitance and V is the potential difference across the capacitor.
Substituting the given values, we get:
W = 1/2 x 2.0 x 10^-6 x (5000)^2
W = 25 J
Therefore, the correct choice is C.

Quiz 24: Capacitance

Doubling the potential difference across a capacitor

A coaxial cable consists of a wire of radius 0.30 mm and an outer conducting shell of radius 1.0 mm.Its capacitance for a 2 m length of the cable is approximately

Which of the following statements about parallel plate capacitor is false?

Two large metallic plates are parallel to each other and charged.The distance between the plates is d.The potential difference between the plates is V.The magnitude of the electric field in the region between the plates and away from the edges is given by

If you increase the charge on a parallel-plate capacitor from 3 µC to 9 µC and increase the plate separation from 1 mm to 3 mm,the energy stored in the capacitor changes by a factor of

Which of the following statements is false?

You want to store 1010 excess electrons on the negative plate of a capacitor at 9.0 V.How large a capacitance must you use?

You make a homemade capacitor out of two flat circular metal plates,each of radius 5 cm,and hold them a distance of 1 cm apart.You then connect each plate to the terminals of a 6-V battery.What would be the capacitance of your capacitor?

You attach a 30-pF capacitor across a 1.5-V battery.How much energy is stored in the capacitor?

A capacitor is constructed by placing a conducting sphere of radius a concentrically inside a thin conducting spherical shell of radius b.Derive an expression for the capacitance of such a capacitor.

A capacitor of capacitance C holds a charge Q when the potential difference across the plates is V.If the charge Q on the plates is doubled to 2Q,

A parallel-plate capacitor has square plates of side 8.0 cm separated by 0.80 mm.If you charge this capacitor to 15 V,the amount of charge transferred from one plate to the other is

Doubling the potential difference across a capacitor

If a capacitor of capacitance 2.0 µF is given a charge of 1.0 mC,the potential difference across the capacitor is

If the area of the plates of a parallel-plate capacitor is doubled,the capacitance is

If the area of the plates of a parallel plate capacitor is halved and the separation between the plates tripled,then by what factor does the capacitance change?

An 80-nF capacitor is charged to a potential of 500 V.How much charge accumulates on each plate of the capacitor?

A coaxial cable consists of a wire of radius 0.30 mm and an outer conducting shell of radius 1.0 mm.Its capacitance per unit length is approximately

If you decrease the charge on a parallel-plate capacitor from 12 µC to 4 µC and increase the plate separation from 1 mm to 3 mm,the energy stored in the capacitor changes by a factor of

A 2.0-µF capacitor has a potential difference of 5000 V.The work done in charging it was

You want to store 10¹⁰ excess electrons on the negative plate of a capacitor at 9.0 V.How large a capacitance must you use?