potential difference capacitance electric current · pdf filepoint a even if there is no test...

Potential Difference

Capacitance

Electric current

ELECTRICITY Lecture 3.8

Electrical Potential Difference

more commonly known as voltage (V)

e.g. 9V battery

Equal height: Equal potential energy

No water flow

Higher potential energy

Lower potential energy

electrons to flow in external circuit

Battery chemical reactions produce electrical potential difference between terminals

Analogy: Open tube filled with water


+ + +

+ + +

+

+ + + +

+

Q

+q

Test charge

F A B

Test charge experiences a repulsive Coulomb force, therefore it has electrical potential energy due to its position

Test charge free to move from A to B, • Electric field does work via Coulomb force • potential energy decreases

∆r

+ +

Definition of voltage

Work done on test charge is where F is the average force on charge +q

ABW F r= ∆

( )B

ABA

W F r dr F r= ≈ ∆∫

For the expert: For point charge, force decreases with distance. For small distances:


Hence potential energy (PE) ∆U of test charge decreases by WAB in going from A to B.

Change in PE, ∆U = -WAB

ABW q∝

U q∆ ∝

ABUVq

∆=

therefore

Voltage between A and B is defined as the change in electric potential energy as charge q moves from A to B

Coulomb force F q∝

UVq

=

V is the potential energy per unit charge

20

14

QqFrπε

=


Voltage is defined as: •potential energy per unit charge •or potential difference

Note: Electric Potential (due to Q) exists at point A even if there is no test charge q there.

SI unit for voltage is the volt (V) 1V = 1J/C

Volt: named after Alessandro Volta (1745-1827), Italian scientist who invented the battery.

Voltage of a battery is the potential difference between its two terminals

V

+ -

+ + +

+ + +

+

+ + + +

+

Q

A

UVq

=

V = - Ed (if the field E is constant)

The energy given to a charge by a voltage is:

Voltage and Electric field

U qV∆ =

Since WAB = Fd,

WAB =-∆U

F=qE (E is the electric field)

Units:

V (volts)

E (NC-1)

d (metres)

WAB = (qE)d = -∆U = -q V } }

Electrical Field (E)

VEd

= −

FEq

=

Volts per metre

Newtons per Coulomb

units are equivalent

Volts Newtonmetre Coulomb

=

1Joule NewtonCoulomb metre Coulomb

=

Joule Newtonmetre

=

Joule Newton metre= ×

UVq

=

Example : The potential at the ground is zero. A storm cloud has a potential of -50kV at an altitude of 500m. What is the associated electric field ?

-50kV

0V

500m

V = -Ed

E = -V/d = -(-50x103)/500 V/m E = 100 V/m

Electric field

Example: In the previous example, calculate the speed of an electron reaching the ground. Electron mass: 9.11x10-31kg.

The potential energy: qV

is transformed into kinetic energy: ½mv²

1/2mv² = qV mqVv 2=

131

319

1011.9)1050)(106.1(2 −

−

−

××−×−

= msv

18103.1 −×= msv

Electric field

For the expert: v~40% of speed of light. Relativistic effects become important!

Electrical capacitance (denoted C) is the ability to store charge, expressed as ratio of charge to potential difference:

where Q is the charge on either plate

Electrical Capacitance

A pair of metal plates separated by an insulator. When subjected to a potential V, charges ±Q will accumulate on the two plates.

-

- - - -

-

+

+ + + +

+

V

Voltage removed: charge remains on the plates

Q V C =

SI unit of capacitance is coulomb per volt: given the name farad (F) (Michael Faraday 1791-1867.- English physicist & chemist)

V

A uniform electric field is created between the plates E=V/d


-

- - - -

-

+

+ + + +

+

- A is the common surface area of the plates - d is their separation - εo= 8.85 x 10-12 C2N-1m-2

is the “permittivity of free space”

Parallel plates separated by free space (≈ air)

0ACdε

=capacitance is given by

Parallel Plate capacitor

QCV

=

d

V


-

- - - -

-

+

+ + + +

+ 0AC

dε

=

Parallel Plate capacitor

Parallel plates separated by free space (≈ air)

-

- - - -

-

+

+ + + +

+

To increase the capacitance insert insulating material between plates

Material referred to as a dielectric

0kACdε

=k known as the dielectric constant

Substance Dielectric Constant Vacuum 1.0 Air 1.00059 paper 3.7 glass

5.6

Example

0kACdε

=

Calculate the capacitance of a parallel plate capacitor of area A =5 cm2 if the plates are separated by a material of thickness d= 0.1 cm and dielectric constant k = 4? If the capacitor is connected to a 9 volt battery, what is the resulting charge on the positive plate? ε0 = 8.85x10-12C2N-1m-2

C = 4 ∗ 5x10-4m2 ∗ 8.85x10-12 C2N-1m-2

0.1 x10-2m

C = 17.6x10-12F = 17.6pF

C = Q/V Q = CV

Q = 17.6 x10-12F *9V = 1.58x10-10 Coulombs

Electrical energy (U) stored in a capacitor


aveU QV=

21

2QUC

=

1

2U QV= 212U CV=

QVC

=

•Defibrillator •Photographic flash unit

Applications

Voltage Applied: Capacitor charges from zero to voltage V,

V

-

- - - -

-

+

+ + + +

+

average voltage (Vave) during charging = ( )0 1

2 2V

V−

=


The plates are moved apart while the battery remains connected.

V

-

- - - -

-

+

+ + + +

+

0ACdε

=

Q CV=

What happens to the capacitance C?

C decreases.

What happens to the potential difference?

V is constant (battery remains connected).

What happens to the charge? Charge Q decreases.


What happens to the charge if battery is disconnected?

V

-

- - - -

-

+

+ + + +

+

0ACdε

=

QVC

=

Charge Q remains constant

Now the plates are moved apart while the battery remains disconnected.

What happens to the charge? Charge Q remains constant. What happens to the capacitance C? C decreases.

What happens to the potential difference? V increases.


Applications:

0kACdε

=

Computer keyboard

Push key •moves plates closer together •Capacitance changes (increases) •detected by computer electronics

S Key

plates d

Random access memory (RAM): Capacitors used in RAM chips to store bits. Touchscreen. A grid of small microscopic capacitors to sense touch.


Defibrillator

Ventricular fibrillation Fast uncoordinated twitching of the heart muscles

Remedy •Strong jolt of electrical energy •Restores regular beating of the heart

Defibrillator •Electrical energy stored in a capacitor •Energy released in ≈ few milliseconds

Electric energy stored in a capacitor Application


Typical Defibrillator (AED)

Defibrillator •Electrical energy stored in a capacitor •Energy released in ≈ few milliseconds

Electric energy stored in capacitor

Magnitude of charge on each plate 6(150 10 )(2250 ) 0.3375Q CV F V C−= = × =

( )( )21 150 2250 3802U F V Jµ= × =

≈200J passed through body in ≈2ms

Therefore power in electrical pulse≈ 100kW

150µF capacitor charged to 2250V

212U CV=


212U CV=

Typical Photographic flash unit

Capacitor (300µF) charged to voltage of 300V

( )21 300 300 13.52U F V Jµ = × =

Energy stored is released rapidly ≈ 10-3 s

Power output of flash unit is ≈13.5 kW

Electric energy stored

Application

The electric current (denoted I) is the charge flowing per second:

It is a measure of the flow rate of charges (analogous to the flow of liquid).

Note:

charged capacitor, charges don’t move, thus I = 0 amps

In electrical cables, charges flow to an appliance when switched on: current I = 0

Electric Current

-

- - - -

-

+

+ + + +

+

SI unit of current is ampere (I) coulomb/second Named after French physicist André Ampere.

electrons

QIt

=

This image cannot currently be displayed.

Electric current not just confined to cables

Electric Current

Lightening strike – large electric field causes air to ionize- large current of very short duration

High dc voltage ≈50kV + -

high energy electrons

x rays

At high electric fields insulators may become conducting: dielelectric breakdown

A defibrillator passes 50 A through the heart for a period of 0.002 s. What is (a) is the quantity of charge passed through the heart in this period? (b) the capacitance of the capacitor supplying this current if it operated at a potential difference of 2000V?

Current =charge /time

Charge q = 50 A x 0.002s = 0.1C

Capacitance =Q/V = 0.1C/2000V= 50µF

Example

qIt

=

Electric Current

Example Current in a torch bulb is 0.2A. How many electrons flow through the bulb if it is on for 5 minutes?

Current =Charge/time

I = Q/t

0.2A = Q/(5*60)

Q = 5*60*0.2 = 60 Coulombs

But charge on the electron =1.6 x10-19C

Therefore number of electrons = 60/(1.6 x10-19) = 37.5x1019 electrons

7. The cell membrane in a nerve cell can be approximated by a parallel plate capacitor with a surface charge density of 5.9 x10-6 Cm-2. Determine the electric field within the membrane. ε0 = 8.85x10-12 C2N-1m -2

QCV

=0ACdε

= VEd

= −

0AQV d

ε= 0 0

VQ A A Ed

ε ε= = −

0

1QEA ε

= −

( )6 25 1

12 2 -1 -2

5.9 106.6 10

8.85 10 C N mCm

E NC− −

−−

×= = ×

×

Example

It requires 5 Joules of energy to move a positive charge of 0.1C from point A to point B. Determine the potential difference between A and B?

+ + q1 q2

+ Q

A B

Potential difference V = energy charge

V = (5 J)/0.1C = 50 Volts

potential difference capacitance electric current · pdf filepoint a even if there is no test...

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