Ch 19 Electric Charges, Forces, and FieldsBackground:

In Ch 7 and 8, we learned about Power = Work / timeo Units à Watts = J / seco P = I Vo J/s = C/s “V”o V = J/C

Electric Charge Electrons reside in a cloud surrounding the

nucleus

These can be ionized, raised or removed from the nucleus.

o When the electrons return to their respective shells, energy emitted in discrete amounts. This allows for identification.

Total electric charge is an universal constant, it is conserved.

o For example is a neutron emits an electron, the neutron become a proton through decay.

Electron's have a negative charge: -1.6 x 10-19 Coulombs Just for fun…me- = 9.11 x 10-31 Kg or

roughly 1/1820 AMUsProton’s are of equal value but positive.

Just for fun…mp+ = 1.67 x 10-27 Kg or about 1 AMU (Atomic Mass Units)

Note: Magnetic Resonance Imaging works on the principle that some atoms can become polarized in a magnetic field, the dipoles can align in the magnetic field.

ExampleA system of 1550 particles, each of which is either an electron or a proton, has a net charge of −6.304×10−17. (a) How many electrons are in this system? (b) What is the mass of this system?

(a) xp+ + xe- = 1550xp+ = 1550 - xe-

xp+ (1.602e-19C)/p++ xe- (-1.602e-19C)/e- = −6.304e−17

(1550 - xe-) (1.602e-19) + xe- (-1.602e-19) = −6.304e−17

xe- = 972 xp+ = 1550 - xe- = 578

(b) 578p+ (1.673e-27kg/p+) + 952 e- (9.11e-31kg/e-) = 9.68 e -25 kg

Insulators and Conductors Conductor: A material whose conduction electrons are free to move throughout. Most metals are

conductors. Insulator: A material whose electrons seldom move from atom to atom. Most insulators are

nonmetals. Excess charge in a conductor is distributed over the surface of the conductor. Semiconductors have properties intermediate between conductors and insulators; their

properties change by doping with other elements. When light shines on a material that becomes conductors, this is said to be Photoconductive.

Coulomb’s Law Like charges à Repel Opposite charged à Attract

q3 = 2q = 2q1 = 2q2

This means q, q1, q2 have equal charge, and q3 has twice the charge.In addition the sketch below implies the distance from q1, q2, and q3 from q are all approximately the same.

When you have multiple point charges, the forces add by superposition

Coulomb's law is the magnitude of the force F between two particles with charges Q and q separated by a distance r.

Coulomb's law |F|=k |q1 q2|/ r2

where k = 1 / 4πϵ0= 8.99 x 109 N m2 / C2

o ϵ0=8.854 x 10−12 C2/(N⋅m2) o is the permittivity of free space

Forces in a Three-Charge SystemExampleConsider two point charges located on the x axis: one charge, q1 = -10.0 nC , is located at x1 = -3.00 m; the second charge, q2 = 40.0 nC , is at the origin (x2 = 0.00 m). What is the net force exerted by these two charges on a third charge q3 = 50.0 nC placed between q1 and q2 at x3 = -1.00 m ? F3 = k [ q1 q3 / (x1-x3)2 - q2 q3 / x32 ]

F3 = −1.91×10−5 N

Comments:Personally I prefer to account for directions by observations, not by math (by math meaning, using the negative and positive signs in the equation) as above.

the negative X1 will pull the positive X3 to the left (thus negative direction) the positive X2 will push the positive X3 to the left (thus negative direction)

Logic and thinking is the most important aspect of Physics. Sadly, my exams conform to accepted standards and logic/thinking is required to a much lesser degree than I prefer.

Electric Field

Electric field is the force per unit charge;

for a point chargeàE = F/q = k |q|/ r2

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Like electric forces, electric fields can also be superposed

The direction of the force depends on the sign of the charge

The electric field of a point charge points radially

positive à away negative à toward

Example An inkjet printer creates a pattern with an electrostatic valve. Let’s assume the ink drops have a mass m = 10 ng each and leave the nozzle and travel horizontally toward the paper at velocity v = 20.0 m/s . The drops pass through a charging unit that gives each drop a positive charge q by stripping electrons. The drops then pass between parallel deflecting plates of length D0 = 2 cm, with a uniform vertical electric field with magnitude E = 100,000 N/C. If a drop is to be deflected a distance d = 0.1 mm by the time it reaches the end of the deflection plate, what magnitude of charge q must be given to the drop?

F = q Em a = q Ea = q E / m (1012 ng = 1 kg)a = q (1e5/10e-12)a = q(1e16)

vave = Δx / Δtvave = D0 / Δt20 = .02 / tt = 0.001 sec

d = ½ (a) t2

1e-4 = ½(qE/m) t2

1e-4 = ½ q(1e16) (.001)2

q = 2 e -14 CoulombsElectric Field Lines

Electric field lines are a convenient way of visualizing the electric field.

Electric field lines:o Point in the direction of the

field vector at every point.o Start at positive charges or

infinity. o End at negative charges or

infinity.o Are denser where the field is

stronger. See Lab for more (below)

For Example in lab In your lab, the most common plate

for drawing in the equipotential lines as well as the perpendicular lines of force is shown to the right.

The equipotential lines are parallel with the conductors.

The lines of force travel from + to – perpendicular to the conductors, thus also the equipotential lines.

à lines of Force are in same direction as the Electric Field

In Lab E = -ΔV / ΔxJust for Fun Example

What is the electric Field in the 10.0 V conductor

E = - ΔV / Δx E = - (10V – 10V) / 2.0 cm E = 0 V/m = 0 N/C

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Shielding and Charging by Induction

Charge resides on outside of conductors (naturally spreads out as much as possible)

Electric Flux and Gauss’ Law Gauss’s law states that the electric flux through a closed surface is proportional to the charge

enclosed by the surface:

In this figure, which figure is easier to calculate the entire flux?Neither, the flux is the same over the entire areaΦ = E A= σ A / ε0 Φ = q / ε0

WHAT is ε0? k(4π) = 1 / ε0ε0 = 1 / k(4π) = 1 / 113x109= 8.85 x 10-12 C2 / N-m2

Example Nerve cells are essentially coaxial cables, or long, thin

cylinders that provide a path of the nerve impulses (electrical signals). The cell membrane of a typical nerve cell consists of an inner and an outer wall separated by a distance of 0.10 μm. The electric field within the cell membrane is 7.0x105 N/C. Treating the cell membrane as a parallel-plate capacitor, what is the magnitude of the charge density on the inner and outer cell walls?SolutionΦ = E A= σ A / ε0 remember à (σ = q / A)σ = Eε0σ = 7.0x105(8.85x10-12)σ = 6.2×10−6 C/m2

Kirchhoff’s 1st LawConservation of electric charge yields that the sum of the current entering any intersection, or node, must equal the sum of any current leaving the intersection.

At Node

n

ΣIk = 0k=1

This is identical for rivers. If three branches meet at one point, branch A at 50 gallons per second, branch B at 200 gallons per second branch C at 150 gallons per second

The resulting larger river leaving from that point must have (50 + 200 + 150) or 400 gallons per second.

Kirchhoff’s 1st Law states that current entering or leaving any junction, or node, must be conserved.i2 = i1 + i3

ExampleE1 = 10 VE2 = 8 VR1 = 20 ΩR2 = 10 ΩR3 = 14 Ω

Kirchhoff’s 2nd LawSince Energy must be conserved, we know that the sum of all voltages must sum to zero in a close loop. This is also referred to as Kirchhoff’s Loop rule.n

ΣVk = 0k=1

Apply 1st LawI2 = I1 + I3

Apply 2nd LawFor top and bottom loops

Top Loop

Bottom Loop

E1 – I1 R1 – I2 R2 = 010 – 20I1 – 10I2 = 010 – 20I1 – 10I1 – 10I3 = 010 – 30I1 – 10I3 = 0

E2 – I3 R3 – I2 R2 + E1 = 08 – 14I3 – 10I2 + 10 = 018 – 14I3 – 10I1 – 10I3 = 018 – 10I1 – 24I3 = 0

Solve system of equationsMultiply bottom loop equation by a factor of -3

10 – 30I1 – 10I3 = 0-54 +30I1 +72I3 = 062I3 = 44I3 = 22/31 = 0.71 ASolve for I1à 10 – 30I1 – 10(.71) = 0I1 = 0.097 A

EMF – Odd term, but let’s relate to your first mechanics class.

If you push an object across a floor with a force of 100 N through 10 meters, how much work have you done?Work = F ΔxWork = 100 N (10 m)Work = 1000 Nm or 1000 Joules

We know from P = I V that a volt has the units of energy per unit charge.

RecapP = I VW = A (V) (Units)J/s = C/s (V)Volt = J / C

So if we have a closed loop and a charge, q, completes the entire loop, the force that caused this charge to move the entire path length this would be described by F Δx per unit charge.

We also know Nm/C is the units of voltage, thus EMF also has the units of Voltage.