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Home , Capacitance, Electric current, Electromotive force

Chapter 19

Electric Potential and the 19.5 and

A plate consists of two plates, one carrying charge +q and the other carrying charge –q.

It is common to fill the region between the plates with an electrically insulating substance called a .

Advantages of dielectrics: • they break down (spark) less easily than air • the capacitor plates can be put closer together • they increase the ability to store charge 19.5 Capacitors and Dielectrics

THE RELATION BETWEEN CHARGE AND POTENTIAL DIFFERENCE FOR A CAPACITOR

The magnitude of the charge in each place of the capacitor is directly proportional to the magnitude of the potential difference between the plates. q = CV

The C is the proportionality constant.

SI Unit of Capacitance: / = (F)

The larger C is, the more charge the capacitor can hold for a given V 19.5 Capacitors and Dielectrics E0

THE DIELECTRIC CONSTANT

If a dielectric is inserted between the plates of a capacitor, the capacitance can increase markedly.

E Dielectric constant κ = o E

Since E ≤ E0 --> κ ≥ 1

Some lines of now terminate or start on induced surface charges

E ≤ E0 E0 E 19.5 Capacitors and Dielectrics 19.5 Capacitors and Dielectrics

THE CAPACITANCE OF A PARALLEL PLATE CAPACITOR

Eo = q (ε o A)

E V E = o = κ d

& κεo A # q = $ !V <--> q = CV % d "

κε A Parallel plate capacitor C = o filled with a dielectric d

C increases for --> increasing κ, increasing A, decreasing d Example. How large would the plates have to be for a 1 F air-filled parallel plate capacitor with plate separation a) d=1 mm, b) d=1 µm ?

a) d = 1 mm

C = κε0A/d = ε0A/d (since κ = 1 for air)

--> A = Cd/ε = (1)(.001)/(8.85 x 10-12) = 1.13 x 108 m2 0 If the plates were square, they would be of and width

(A)1/2 = (1.13 x 108)1/2 = 11 km --> very large!!

b) d = 1µm = 10-6 m ~ the thickness of paper – use paper: κ = 3.3

-6 -12 4 2 --> A = Cd/(κε0) = (1)(10 )/(3.3 x 8.85 x 10 ) = 3.42 x 10 m

and (A)1/2 = (3.42 x 104)1/2 = 185 m --> about 2 football fields in length, improving! 19.5 Capacitors and Dielectrics

Conceptual Example 11 The Effect of a Dielectric When a Capacitor Has a Constant Charge

An empty capacitor is connected to a and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material is inserted between the plates. Does the across the plates increase, remain the same, or decrease? Effect of a Dielectric When a Capacitor Has a Constant Charge -- continued

Before inserting dielectric:

q = C0 V0, C0 = ε0A/d

After inserting dielectric:

q = C V, C = κε0A/d = κC0

C0 V0 = q = C V = κC0 V

--> V = V0/κ --> V < V0 for κ > 1

i.e. the voltage across the plates decreases when the dielectric is inserted if charge is held constant 19.5 Capacitors and Dielectrics

ENERGY STORAGE IN A CAPACITOR Capacitors store charge at some potential difference, therefore they store EPE, since by definition for a point charge q, EPE = qV . Your book shows that the energy stored in a capacitor can be written several ways:

Energy = 1/2 qV = 1/2 CV2 = q2/(2C) --> useful forms to solve problems

2 Using Energy = 1/2 CV , C = κε0A/d, and V = Ed , the energy of a capacitor can be written as, Volume = Ad

1 & κεo A # 2 Energy = 2 $ !(Ed ) % d "

We can then calculate the energy density of a capacitor as,

Energy 1 2 (valid in general for the Energy density = Volume = 2 κεo E energy density stored in an electric ) Example. How much energy is stored in a 10 F capacitor with a potential difference of 3 V across it?

Energy = 1/2 CV2 = 1/2(10)(3)2 = 45 J

This is the same amount of energy stored in a 4.6 kg suspended 1 m above the . Chapter 20

Electric Circuits 20.1 and Current

In an electric circuit, an energy source and an energy consuming device are connected by conducting through which electric charges move. 20.1 Electromotive Force and Current

Within a battery, a chemical reaction occurs that transfers from one terminal to another terminal.

The maximum potential difference across the terminals is called the electromotive force (emf).

The emf has units of , e.g. a 12 V battery has an emf of 12 V 20.1 Electromotive Force and Current

The is the amount of charge per unit that passes through a surface that is perpendicular to the motion of the charges.

Δq I = Δt

One coulomb per equals one (A). 20.1 Electromotive Force and Current

If the charges move around the circuit in the same direction at all , the current is said to be (dc) --> e.g. simple circuits with batteries are normally dc.

If the charges move first one way and then the opposite way, the current is said to be (ac) --> e.g. the current coming out of your electric outlet is ac.

20.1 Electromotive Force and Current

Example 1 A Pocket

The current in a 3.0 V battery of a pocket calculator is 0.17 mA. In one hour of operation, (a) how much charge flows in the circuit and (b) how much energy does the battery deliver to the calculator circuit?

(a) Δq = I(Δt)= (0.17×10−3 A)(3600 s)= 0.61 C

ΔV = -W/q

Energy Energy = Charge × = (0.61 C)(3.0 V)=1.8 J (b) Charge 20.1 Electromotive Force and Current

Conventional current is the hypothetical flow of positive charges that would have the same effect in the circuit as the movement of negative charges that actually does occur. 20.2 ’s Law

It is found experimentally that the current flowing through a circuit is proportional to the voltage across the circuit, i.e.,

I is proportional to V

The resistance (R) is defined as the ratio of the voltage V applied across a piece of material to the current I through the material. 20.2 Ohm’s Law

OHM’S LAW

The ratio V/I is a constant, where V is the voltage applied across a piece of material and I is the current through the material:

V = R = constant or V = IR I Resistance

SI Unit of Resistance:

volt/ampere (V/A) = ohm (Ω) 20.2 Ohm’s Law

Resistance, R, represents to what extent the current can flow freely in the circuit, i.e. the larger R, the more the electrons scatter with in the material.

These scatterings slow down electrons and transfer energy as to the material.

To the extent that a or an electrical device offers resistance to electrical flow, it is called a . 20.2 Ohm’s Law

Example 2 A Flashlight

The filament in a bulb is a resistor in the form of a thin piece of wire. The wire becomes hot enough to emit light because of the current in it. The flashlight uses two 1.5-V batteries to provide a current of 0.40 A in the filament. Determine the resistance of the glowing filament.

V 3.0 V R = = = 7.5 Ω I 0.40 A