Electric Charge & Electric Fields

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Electric Charge & Electric Fields

Objects with excess + or - charge give rise to electric force.

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Acquisition of Charge

Solid objects charged by e- transfer.+charge results from loss of e--charge fr gain of e-.

Liquids or gasses + and - ions are free to move about.

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Neutral objects can be polarized

Charges which are free to move are redistributed.

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Charged objects can induce polarization. Balloon is attracted to positive wall surface.

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Outer Part Q Elem Qelectrons e- –1 -1.6 x 10-19 C

NucleusProtons p+ +1 1.6 x 10-19 Cneutrons no 0 0

Quantity of Charge in Atoms

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Conservation of Charge Q

Q can be transferred, Q cannot be created or destroyed.

Q in system remains constant.

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Solids.

• Conductors – allow charges to move around (metals).

• Insulators – hold excess charge in place.

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Conductor distribute Q.

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Insulators – charges concentrated in one spot.

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Polarization occurs easily in conductor.

• Friction – rub 2 objects. Insulators.

• Conduction (contact) – touch charged to uncharged q shared equally. Conductors.

• Induction – polarization & grounding (drives off or sucks in e-). 1 conductor 1 insulator.

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Charging Objects:

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Conductors can be charged or discharged by induction. Need to ground (Earth).

1. A positively charged glass wand with a charge of +1000 is used to charge a neutral metal sphere by induction. The resulting charge on the sphere will be:

• 1) + 500

• 2) - 500

• 3) +1000

• 4) -1000.

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Charge q is quantized. There is a smallest unit.

Charge can only exist in whole number integers of the charge on e.

Conservation of Charge

A. How many electrons does it take to carry a charge of -1.0 C?

Units of charge = coulombs (C) Charge on e- is -1.6 x 10-19 C

Charge on p+ is +1.6 x 10-19 C

fundamental / elementary units, e:e- has charge –1p+ has charge +1

6.25 x 1018

2. Which Charge cannot exist on an object?

• 1) 5.6 x 10-19 C.

• 2) 1.15 x 1015 e.

• 3) 1.12 x 10-18 C.

• 4) 1 billion billion e.

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Coulomb’s Law Relates Force btw. charged objects.

Fe = kq1q2 r2

k = constant 8.99 x 109 N m2/C2. q charge on obj in Coulombs (C)r is dist btw centers meters.F is force (N)

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The constant k can be written as:

k = 1/4.

Where is the permittivity of free space in a vacuum (air)= 8.85 x 10-12 N m2/C2.

3: State Coulomb’s Law in words.

• The force between 2 point charges is directly proportional to the amount of charge and

• Inversely proportional to the square of the distance between their centers.

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Ex 4 . Charges (3) of +1 C are located at the corners of a 45o right triangle. What is the resultant force on the charge located at the 90o angle.

1m

1m

1C1C

1C

1.3 x 1010 N45o below horz

F1

F2

Electric Force is vector quantity.

Hwk Rd Kerr 6.2.1 – 6.2.4do pg 162 #1 – 3, 7, 9,12, 14, 17 .

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Electric Fieldregion of space around charged object where a charge feels an

electrostatic force.

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Electric Fields-Charge alters space around it. Charged objects feel a force.

Either repulsion or attraction.

Electric Field (E) defined as:

The force per unit charge at a point in space on a small +test charge..

E = F/q.

E = Electric Field (N/C)F is force on test charge (N).q is amt of charge on test charge (C).

Remember gravitational field?

•Region of space where mass feels a force.

•g = force/unit mass on a small mass N/kg or m/s2.

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5. Calculate the E field strength 0.4 m away from a charge +20 C.

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E = F/q

Fe = kQ1q2

r2.Sub in for F kQ1q2 where q = q2.

r2

q

E = kQ1 Memorize .r2

Fill in the numbers.

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E = kQ1

r2

E = 9x109(20x10-6)C

(0.4)2

1.1 x 106 N/C.

Field Lines represent electric fields.

Electric field lines show the force that a small positive test charge feels in a field created by a much larger charge. They represent the strength and direction of the field.

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Sketch vectors to show force magnitude & direction on a + test charge at each point.

+

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Field around positive object. Lines start on + charge, end on –

charge.

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The denser the field lines are, the stronger the field.

Stronger field near charge.

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Field lines start on + and end on – charge.

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Field Between Parallel Plates

How would the strength of the field vary if a charge moves from the + to the – plate?

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Fields have strength and direction - vector.

Density of lines shows strength.Direction shown as arrows.

Direction is determined by a + test charge.Lines start on pos end on neg.

Electric Field lines don’t touch or cross.

Electric field due to more than one charge.

Field is stronger near the larger charge. Density of lines show the increased strength.

Strength at a point is the vector sum of field strengths.

Electrostatic Equilibrium

Fields produced by more that a single charge will have spots where the forces on a charge in the field will be balanced.

F net = 0.

E field inside a conductor is zero!

• Why?

• If an E field existed inside a conductor, and q are free to move, the E field would exert Fnet on all q present. Fnet would accelerate the q until equilibrium were reached and E goes to zero.

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Superposition

To find the force or field on a charge q, due to the presence of more than one other charge, you must use vector addition.

6: Two 10 C charges separated by 30 cm. What is the field strength 10 cm to the right of A?

• 6.75 x 10 6 N/C

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10 C 10 C

10 cm 20 cm

• B

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1909 Millikan measured charge on e-

Drops suspended when Fg = Fe.

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Hwk Kerr pg 162 #4 – 6, 8 10, 11, 15 16, 18 -21, 23

Film Mech Universe E fields

Work & Energy

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Charges in an E field can have PEelc.

It takes work to bring charges near a repelling charge causing charge to gain PE.

Charges that are attracted by opposite charges are said to “fall” toward them losing PE. Work is done by the field.

The convention:

• If the charge gains PE then work is positive.

• If it loses PE work is negative.

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Where will a charge feel no force from electric field?

Infinity!

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Electric Potential

• Wk per Coulomb to bring a charged particle to point in field from infinity.

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It requires energy to bring +q fr. infinity to point P in E field. They repel. Each point in a field has electric potential, like a height.

+q

P

Must put a force on +q & push it charge thru a distance.

Electric Potential

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The amt of work done on every coulomb of charge moving it is called electric potential, V.

V = W/q. W work in Jq is charge in C.V is Volts = J/C.

V defines the potential at P also called voltage.P is like a particular height in a gravity field.

Note: Wk also = E.

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Note: the work done by an electric field is conservative!

It is independent of the path taken and depends only on the endpoints.

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Since work is done to move charge, the PEelc of q must change:

V = W/q but W = PE.

W = PE =qV.

Potential DifferenceImagine moving a 2 C charge from A to B in field where A and B are at dif potentials (heights). What is the dif in potential?

A = 15 VB = 28 V

pd = 28V – 15V = 13 V.

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In Uniform Field

Constants between plates:F on q.field intensity E= F/q, work done (Fd) to move a charge

potential difference, V.

V = W V = Fd F = E

q q q

V = Ed

E = V/d

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Prove that for uniform field: E = V

d

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Since: E = V d

for a constant Electric field

Units of E can be V/m

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Energy of Moving Charges in Fields.

• As a charge moves thru a field, its total E is constant. By consv of E:

• If a charges “falls” toward the oppositely charged plate its PEelc decreases, but it will accelerate, its KE increases.

• Work done by field will accelerate charge W = KE = qV.

• So: before after

• ET = ET .

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There is a potential difference – voltage between the two plates based on their charge & distance between them.

A proton near the positive plate is at a high potential (energy).

What is the potential of a proton stuck to the negative plate?

(0)

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The electron-volt: a unit of work & energy.

For very small changes in PEelc (on the order of 10-19J) eV is used.

The electron-volt, eV, is the work & E required to push 1 e- (or p+) through a voltage of 1V.

W = qV = (1.6 x 10-19 C)(1V) = 1.6 x 10-19 J = eV.

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If 1 e- is pushed across 1V then 1 eV of work is done.

If a charge of 2e- is pushed across a 1V pd then work = 2 eV.

If 2e- pushed across 6V then work is 12 eV.

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36 eV

What if 3e- move across 12 V?

To find eV (# elm charges) (voltage)

7. How many joules of energy are represented by 6.9 x 1029 eV.

6.9 x 1029 eV x 1. 6 x 10-19 J. = 1.1 x 1011 J

eV

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Review: Voltage or Electric Potential

Wk per Coulomb to bring a charged particle to point in field from infinity.

Potential / Voltage difference Wk per Coulomb to move charge between two points at different potentials.

Charges in field have PEelc.High PE charge near point with same charge.Low PE charge near point with opposite charge.

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Charges set loose in E fields will accelerate!

By conservation E

PE lost = KE gained.

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Some typical voltages

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Plates with battery

AC Delco 12 volts

d

- +d = 1 cm

A B

Batteries are meant to maintain the constant potential difference & electric field.

ABV Ed

AB /E V d12 / 0.01E 12000 N/CE

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Mech Uni Potential Dif

PE elc = KE

qV = ½ mv2.

If a p+ is released from rest near the positive plate in a uniform field, it will accelerate toward the negative plate changing PE to KE so:

A proton is released from rest in a pd = 1200 V. What will be its maximum speed?

v = 4.8 x 10 5 m/s

PE elc = KE

qV = ½ mv2.

(1.6 x 10-19 C)(1200 J/C) = ½ (1.67x10-27kg)(v)2.