ECE 1100: Introduction toECE 1100: Introduction toElectrical and Computer EngineeringElectrical and Computer Engineering

Notes 17

Electric Field, Voltage, and Power

Spring 2008

David R. JacksonProfessor, ECE Dept.

Electric FieldElectric Field

+

-V0

+ + + + + + + + + + + + + + + + + + + +

- - - - - - - - - - - - - - - - - - - - - - - - - - - -

x

A

B

hq (“test charge”)Fx

Fx = force on test charge (proportional to q)

/x xE F q The electric field strength is the force per unit charge

Electric Field (cont.)Electric Field (cont.)

x xF q E

For a “unit” test charge (q = 1 [C]):

x xF E

We can also write this as

Electric Field (cont.)Electric Field (cont.)

The electric field vector:

The direction of the vector is the direction of the force on a q = 1 [C] test charge.

The magnitude of the vector is the magnitude of the force on a q = 1 [C] test charge.

F qEIn general,

For a “unit” test charge (q = 1 [C]) : F E

Electric Field (cont.)Electric Field (cont.)

Inside a parallel-plate capacitor, the electric field vector is (approximately) a constant.

Parallel-plate capacitor

+

-V0

+ + + + + + + + + + + + + + + + + + + +

- - - - - - - - - - - - - - - - - - - - - - - - - - - -

x

A

B

hE

Electric Field (cont.)Electric Field (cont.)

The electric field vector is pointed in the radial direction in spherical coordinates, and decreases with distance from the point charge.

q

Point charge

Voltage DropVoltage Drop

VAB voltage (at A) – voltage (at B)

VAB Ex h

Note: the unit of electric field is volts/meter

+

-V0 x

A

B

h

ˆxE xE

- - - - - - - - - - - - - - - - - - - - - - - - - - - -

+ + + + + + + + + + + + + + + + + + + +

Voltage DropVoltage Drop

VAB Ex h

Notes:

The electric field vector points from the positive charges to the negative charges. The electric field vector points from the higher voltage to the lower voltage.

+

-V0 x

A

B

h- - - - - - - - - - - - - - - - - - - - - - - - - - - -

+ + + + + + + + + + + + + + + + + + + +

Voltage Symbol (Reference Direction)Voltage Symbol (Reference Direction)

V voltage (at + sign) – voltage (at – sign)

Voltage drop symbol (reference direction):

+ -V

Note: V may be numerically positive or negative.

Given: V = -3 [V]

ExampleExample

Find: VAB

+ -V

B A

VBA = V(B) – V(A) = V = -3 [V]

VAB = -VBA = - (-3)

VAB = 3 [V]

Voltage Source: BatteryVoltage Source: Battery

VAB = V(A) – V(B) = V0

Battery (constant voltage)

V0 = battery voltage

A

B

V0 [V]

anode (+ terminal)

cathode (- terminal)

Note: V0 is always positive for a battery.

Voltage SourceVoltage Source

Time-varying voltage source

vAB = vA (t) – vB (t) = vs (t)

+-vs (t)

A

B

Note: A is not always at the higher voltage! (i.e., vs(t) might be

negative for some value of time).

Example (Find the Voltage)Example (Find the Voltage)

V = VAB = +9 [V]

A

9 [V]

B

+

-V

V = VAB = -VBA = -9 [V]

B

9 [V]

A+

-

V

+-3sin(t) [V]

+

-

v

v = 3sin(t) [V]

+-3sin(t) [V]

-

+

v

v = -3sin(t) [V]

Example (Find Example (Find VV))

V = VAB = - VBA = -9 [V]

9 [V] 9

+

-

V

A

B

Note: there is no loss of voltage across an ideal wire (zero resistance).

Relation Between Voltage and WorkRelation Between Voltage and Work

x x x ABW F h qE h q E h qV

+

-V0 x

A

B

hq (“test charge”)Fx

The test charge is moved from A to B:

W = energy (work) given to particle by the electric field

Relation Between Voltage and Work (cont.)Relation Between Voltage and Work (cont.)

/ABV W q

The voltage drop between two points is the energy given to a unit test charge (q = 1 [C]) that moves between the two points.

ABW qV

or

ExampleExampleAn electron moves from the negative to positive terminal of a 12 [V] battery (starting from rest). How much speed does it pick up?

21

2ABW qV mv 2 ABqVv

m

- +

A B

or

q = - 1.6022 10 -19 [C]

VAB = - 12 [V]

m = 9.1091 10 -31 [kG]

v = 2.055 106 [m/s]

Power DissipationPower Dissipation

device

i

+ -v

Assume a “Passive Sign Convention” :

The reference direction for current points through the device from + to - .

Let Pabs = power absorbed by device

Power Dissipation (cont.)Power Dissipation (cont.)

In time t: q i t

AB ABW q v i t v i t v vi t

/absP W t vi

device

i+ -v

q

A B

Note: For a resistor, the energy given to the charges eventually ends up as heat.

Let W = energy given to charge q as it moves from A to B.

Power Dissipation (cont.)Power Dissipation (cont.)

absP vi

Note: the power delivered (provided) by a device is the negative of the power absorbed by the device.

del absP P

ExampleExample

Find Pabs by resistor

9[V] 1[A] 9 [W]absP VI

9 [W]absP

1 [A]

+

-9 V 9

+

-

V

Example (done a different way)Example (done a different way)

Find Pabs by resistor

9[V] 1[A] 9 [W]absP VI

9 [W]absP

+

-9 V 9

-1 [A]

+

-

V = -9 [V]

Example Example

Find Pabs by battery

9[V] 1[A] 9 [W]absP VI

9[W]delP 9[W]absP or

+

-9 V 9

-1 [A]

Example Example

Find Pabs by each element

3 [A]

+ -2 [V]

(2)(3) 6 [W]absP VI

3 [A]

+ -2 [V]

(2)(-3) 6 [W]absP VI

Example (cont.) Example (cont.)

Find Pabs by each element

2 [A]

+ --1 [V]

( 1)( 2) 2 [W]absP VI - -

- 2 [A]

+- 1 [V]

(1)(2) 2 [W]absP VI

Example Example Find Pabs (t) and the energy Wabs, defined as the energy that is

absorbed by the device for 3 < t < 4 [s].

+

-

vx (t)

ix (t) 12 [V]xv t

5 [A]txi t e

12 5 60t tabs x xP t v t i t v t i t e e

60 [W]tabsP t e

Given:

Example (cont.) Example (cont.)

+

-

vx (t)

ix (t) 60 [W]tabsP t e

absabs

dWP t

dt

4 4

3 3

4 3absabs abs abs abs

dWP t dt dt W W W

dt

The energy absorbed in the time interval (3,4) is:

Example (cont.) Example (cont.)

4

360 t

absW e

1 888 [J]absW .

+

-

vx (t)

ix (t)

4 4

3 3

4

3

60

60

tabs abs

t

W P t dt e dt

e