front plate slightly chargedinduces opposite charge on back plate.

Brushes pull off chargescharges collected in leyden jar (capacitor)W

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urs

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Electric Field

Definition

• Electric field is the

strength & direction of the forces in space surrounding a positive test charge

Electric field as coulomb’s law

• F = q1 * (k q2 /d2)

rewrite the force equation

F = q1 * E

q1 is the positive test charge

E is the field created by charge(s) q2

Top tips on electric fieldE=0 inside a metal (faraday) cageField lines go from + to – charges

Closer field lines are stronger

Electric Field between 2 unlike charges

+ -

A

B

C

Note that the distance betweenelectric field lines at C is shorterthan that at B.

Distance between electric fieldlines at B is shorter than at A.

Electric Field between 2 charged plates

positive

negative

Electric Field between 2 charged plates

+ + + + + + + +

- - - - - - - -

coulombs, joules, volts

• Coulomb= unit of charge

• Joule = unit of energy

• Volt= unit of energy per charge

coulomb• How many charges are in 1C? coulomb?• 1C is about 6 billion billion charges)

(1/1.6 x 10-19 = 6.25 x 1018)• The man, the law

Charles Coulomb~1750

Joules of energy

• Charges have a type of energy called

electron potential energy (PEe or U)

High energy: + is close to +: ++

or + is far from –: + -

It takes work to move charges against (opposite) an electric field High energy work needed to move + charge

(W = F*d)

E

F

+

low energy

POSITIVE charge moving in an E field.

(a) When a positive charge moves in the direction of an electric field, the field does positive work

and the potential energy decreases. Work = qo

E d

Volts is an energy density

• Voltage is also called potential• 1 volt = 1 joule / 1 coulomb• Example: 12 V battery: every coulomb of

charge has 12 joules of energy

Examples

• Static balloon 9 V battery (1hr,1A)

1 joule of energy 9x104 joule energy

0.001 C of charge 1x104 C of charge

1/.0001 = 1000 volts 9/1 = 9 Volts

Another c,j,v example

• Van de graaf static generator has..

1,000,00 volts (high) 1 joule of energy (low) .00001 coulombs of charge

1,000,000 v = 1 j/ .00001c

Uniform electric field 2 situations Point Charges

Force: F=q*E Force= k q1*q2 / d2

Electric potential energy Electric potential energy

PE= F*d = q*E*d PE = F*d = k q1*q2 / d

Voltage (electric potential) Voltage (electric potential)

V= PE/q1 = q*E*d /q = E*d V= PE/q1 = k q2 / d2

Voltage difference: Voltage difference:

DVe = D PE/q1 = E* Dd DVe = D PE/q1 = k q2 / d2 - kq2/ d1

It takes work to move + charge against static forces like it takes work to lift against gravity

• PEfinal

• +++++

• +

• Peiniitial work = gain in PE = F*d

work

Electrostatics gravitational

Force: F= kq1 q2 / d2 F= Gm1m2/d2

Field E= F/q1 = kq2/d2 g = F/m1 = Gm1/d2

potential energy PE= F*d = kq1 q2 / d PE = F*d= Gm1m2/d

(using field) =qE*d = mg*h

Potential V= PE/q1 = kq2 / d U= PE/m1 = G m2 / d

(Using field) = E*d = g*h