26 Ch.

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Mid Exam

Monday 07:00-08:30 PM 02/03/1442H

19/10/2020G

Capacitance and Dielectrics

Ch. 23

Phys 104 - Ch. 26/I - lecture 12 1442 - 1

^st

semester

Dr. Ayman Alismail

Ch. 26

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26.01

sec.

➢ Acapacitor consists of two conductors carrying charges of equal magnitude and opposite sign.

➢ A potential difference ∆𝑉 exists between the conductors due to the presence of the charges.

➢ The quantity of charge 𝑄 on a capacitor is linearly proportional to the potential difference∆𝑉between the conductors:

𝑄 ∝ ∆𝑉

➢ The proportionality constant 𝐶(known as the capacitance) depends on the shape and separation of the conductors. We can write this relationship as:𝑄 = 𝐶∆𝑉.

➢ The capacitance 𝐶 of any capacitor is the ratio of the charge 𝑄 on either conductor to the potential difference∆𝑉 between the conductors:

𝐶 = 𝑄

∆𝑉

➢ The capacitance is always a positive quantity.

➢ The capacitance is a measure of a capacitor’s ability to store charge. Because positive and negative charges are separated in the system of two conductors in a capacitor, there is electric potential energy stored in the system.

➢ The SI unit of capacitance is coulombs per volt, or thefarad (F),1 F = 1 ΤC V.

All of these devices are capacitors, which store electric charge and energy. A capacitor is one type of circuit element that we can combine with others to make electric circuits.

Definition of Capacitance

A capacitor consists of two conductors. When the capacitor is charged, the conductors carry charges of equal magnitude and opposite sign.

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26.01

sec. Definition of Capacitance

A parallel-plate capacitor consists of two parallel conducting plates, each of area 𝐴, separated by a distance 𝑑. When the capacitor is charged by connecting the plates to the terminals of a battery, the plates carry equal amounts of charge. One plate carries positive charge, and the other carries negative charge.

➢ Let us consider a capacitor formed from a pair of parallel plates.

▪ Each plate is connected to one terminal of a battery, which acts as a source of potential difference.

▪ If the capacitor is initially uncharged, the battery establishes an electric field in the connecting wires when the connections are made.

▪ On the plate connected to the negative terminal of the battery, the electric field applies a force on electrons in the wire just outside this plate; this force causes the electrons to move onto the plate.

▪ This movement continues until the plate, the wire, and the terminal are all at the same electric potential.

▪ Once this equilibrium point is attained, a potential difference no longer exists between the terminal and the plate, and as a result no electric field is present in the wire, and the movement of electrons stops. The plate now carries a negative charge.

▪ A similar process occurs at the other capacitor plate, with electrons moving from the plate to the wire, leaving the plate positively charged.

▪ In this final configuration, the potential difference across the capacitor plates is the same as that between the terminals of the battery.

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26.01

sec.

Problem 26.01

(a) How much charge is on each plate of a 4.00 μFcapacitor when it is connected to a 12.0 V battery? (b) If this same capacitor is connected to a 1.50 V battery, what charge is stored?

𝑄 = 𝐶∆𝑉 = 4.00 × 10

⁻⁶

× 12.0 = 4.80 × 10

⁻⁵

C = 48.0 μC 𝑄 = 𝐶∆𝑉 = 4.00 × 10

⁻⁶

× 1.50 = 6.00 × 10

⁻⁶

C = 6.00 μC

Definition of Capacitance

(a)

(b)

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26.02

sec. Calculating Capacitance

➢ The capacitance depends only on the geometry of the conductors and not on an external source of charge or potential difference.

➢ There are three familiar geometries of the capacitors, namely, parallel plates, concentric cylinders, and concentric spheres.

Parallel-Plate Capacitors

▪ Two parallel metallic plates of equal area 𝐴 are separated by a distance 𝑑 (assuming that the charged conductors are separated by a vacuum).

▪ One plate carries a charge𝑄 , and the other carries a charge−𝑄.

▪ The amount of charge that can be stored on a plate for a given potential difference increases as the plate area is increased. Thus, we expect the capacitance to be proportional to the plate area𝐴.

▪ Moving the plates closer together increases the potential difference between the plates and causes the charge on the capacitor to increase. If 𝑑 is increased, the charge decreases. As a result, we expect the capacitance of the pair of plates to be inversely proportional to𝑑.

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26.02

sec. Calculating Capacitance

➢ The value of the electric field between the plates is:

𝐸 = 𝜎

𝜖

₀

= 𝑄 𝜖

₀

𝐴

Because the field between the plates is uniform, the magnitude of the potential difference between the plates equals𝐸𝑑; therefore:

∆𝑉 = 𝐸𝑑 = 𝑄𝑑 𝜖

₀

𝐴

➢ We find that the capacitance is:

𝐶 = 𝑄

∆𝑉 = 𝑄 Τ 𝑄𝑑 𝜖

₀

𝐴 𝐶 = 𝜖

₀

𝐴

𝑑

➢ That is, the capacitance of a parallel-plate capacitor is proportional to the area of its plates and inversely proportional to the plate separation.

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26.02

sec. Calculating Capacitance

➢ The electric field is uniform in the central region between the plates.

However, the field is nonuniform at the edges of the plates. Such edge effects can be neglected if the plate separation is small compared with the length of the plates.

The electric field between the plates of a parallel-plate capacitor is uniform near the center but nonuniform near the edges. (b) Electric field pattern of two oppositely charged conducting parallel plates. Small pieces of thread on an oil surface align with the electric field.

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26.02

sec. Calculating Capacitance

➢ Let us assume a battery connected to a single parallel-plate capacitor with a switch in the circuit.

▪ When the switch is closed, the battery establishes an electric field in the wires and charges flow between the wires and the capacitor. As this occurs, there is a transformation of energy within the circuit.

▪ Before the switch is closed, energy is stored as chemical energy in the battery.

▪ This energy is transformed during the chemical reaction that occurs within the battery when it is operating in an electric circuit.

▪ When the switch is closed, some of the chemical energy in the battery is converted to electric potential energy related to the separation of positive and negative charges on the plates.

▪ As a result, we can describe a capacitor as a device that stores

energy as well as charge. (a) A circuit consisting of a capacitor, a battery, and a switch. (b) When the switch is closed, the battery establishes an electric field in the wire that causes electrons to move from the left plate into the wire and into the right plate from the wire. As a result, a separation of charge exists on the plates, which represents an increase in electric potential energy of the system of the circuit. This energy in the system has been transformed from chemical energy in the battery.

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26.02

sec.

Example 26.01

A parallel-plate capacitor with air between the plates has an area𝐴 = 2.00 × 10⁻⁴ m² and a plate separation𝑑 = 1.00 mm. Find its capacitance.

𝐶 = 𝜖

₀

𝐴 𝑑

𝐶 = 8.85 × 10

⁻¹²

× 2.00 × 10

⁻⁴

1.00 × 10

⁻³

𝐶 = 1.77 × 10

⁻¹²

F 𝐶 = 1.77 pF

Calculating Capacitance

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26.02

sec.

Problem 26.07

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm², separated by a distance of 1.80 mm. A20.0 V potential difference is applied to these plates. Calculate (a) the electric field between the plates, (b) the surface charge density, (c) the capacitance, and (d) the charge on each plate.

Calculating Capacitance

∆𝑉 = 𝐸𝑑 𝐸 = ∆𝑉

𝑑 = 20.0

1.80 × 10

⁻³

= 1.11 × 10

⁴

V m = 11.1 Τ kV m Τ 𝐸 = 𝜎

𝜖

₀

𝜎 = 𝜖

₀

𝐸 = 8.85 × 10

⁻¹²

× 1.11 × 10

⁴

= 9.83 × 10

⁻⁸

C m Τ

²

= 98.3 nC m Τ

²

𝐶 = 𝜖

₀

𝐴

𝑑 = 8.85 × 10

⁻¹²

× 7.60 × 10

⁻⁴

1.80 × 10

⁻³

= 3.74 × 10

⁻¹²

F = 3.74 pF

∆𝑉 = 𝑄 𝐶

𝑄 = 𝐶∆𝑉 = 3.74 × 10

⁻¹²

× 20.0 = 7.47 × 10

⁻¹¹

C = 74.7 pC

(a)

(b)

(c)

(d)

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26.02

sec.

Problem 26.09

When a potential difference of 150 Vis applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of30.0 nC cmΤ ² . What is the spacing between the plates?

𝑄 = 𝜖

₀

𝐴 𝑑 ∆𝑉 𝑄

𝐴 = 𝜎 = 𝜖

₀

𝑑 ∆𝑉 𝑑 = 𝜖

₀

𝜎 ∆𝑉

𝑑 = 8.85 × 10

⁻¹²

3.00 × 10

⁻⁴

× 150 = 4.42 × 10

⁻⁶

m = 4.42 μm