CHAPTER 17CHAPTER 17Electrical Energy and CurrentElectrical Energy and Current

Conservative Forces:Conservative Forces: Work done on an object depends only on its initial and final position. The path from the initial to the final position is not important

ExampleExample (Gravitational Force)Consider the work done on an object against gravity. W = Fg d = mghWork is converted to gravitational potential energyThe Electrostatic Force (Fe) is also a conservative force.

(qE = = Fe)k q1 q

r2W = Fe d = qEd

Work done by gravity (falling object) = -mgh

Work done by electrostatic force = -qEd

W = q E d

Potential Difference:Potential Difference: The change in potential energy (Electric Potential) of a charge, q1 divided by q1.Potential DifferencePotential Difference Change in potential energyPotential Difference (V)(Electric Potential)

• PE occurs in a uniform electric field• q is a charge that changes position in the uniform field• V is a scalar quantity• V units = Joules/Coulomb 1 Volt (V) = 1 Joule/Coulomb

V = PE q

Example ProblemExample Problem (moving a positive charge against an electric field)

q

15cm E = 250 N/Cq = +400Cd = 15cm

Work done on the charge = qEd= (400x10-6C)(250N/C)(15x10-2m)= 1.5 x 10-2 Joules

Work done increases the potential energy of the charge.

V = PE q

= qEd q

= Ed

V = (250N/C)(15x10-2m)V = 38 Nm/CV = 38 J/C

V = 38VNOTE: [E] = V/m = N/C

SummarySummary

Sign of ChargeSign of ChargeDirection of Movement Direction of Movement

Relative to E-FieldRelative to E-Field Sign of Sign of PEPE

+

–

+

–

Opposite (against)

Opposite (against)

Same direction (with)

Same direction (with)

+

–

–

+

Electric Field Between Parallel PlatesElectric Field Between Parallel PlatesWhat is the Electric Potential Differenced in the above diagram if E=25x102N/C and d=15cm?

Electric Potential Difference = V = VB – VA = -Ed(Potential Difference)

= V = -(25x102 N/C)(15x10-2 m) = V = -375 Nm/C = V = -375 J/C

V = -375V

If the charge consisted of a proton:

m = 1.67x10-27kg q=1.60x10-19C

V= 38J/CWhat would be its change in potential energy and with what velocity would it be moving at “B” if it was at rest at point “A”?

StrategyStrategyCalculate change in potential energy and convert potential energy to kinetic energy. Solve for v.PE = qV = (1.60x10-19C)(38J/C)PE = 6.1x10-19JPE = KE = KEB (KEA = 0)

6.1x10-18J = (1.67x10-27kg) v2

2v = 6.0x104 m/s

Electric Potential Associated with Point ChargesElectric Potential Associated with Point ChargesBetween parallel plates E is uniform.E associated with a point charge is not uniform

E = k q r2

= k q d2

r = d

V = Ed for only small d valuesCalculus to the rescue!

dv = Edd

dv = ddk q d2

dv = dd

k q d2

d= r

d=

v = -1 k qd

d=r

d=

v= k qr

v = k q

rScalar Quantity

Electric Potential Caused by Point Change

Electric Potential Caused by 2 Point ChargesElectric Potential Caused by 2 Point ChargesWhen Analyzing Multiple Point Charges:When Analyzing Multiple Point Charges:

• The principle of superposition applies. (Just like with calculating E-field due to multiple charges.)• However “v” is a scalar quantity (J/C) and “E” was a vector quantity (N/C)• Scalars are much easier to add than vectors because with scalars… we have no direction.

Example ProblemExample Problem (Electric Potential : linear)Two point charges 20cm apart each with a charge of +50C are established. What is the electric potential 10.cm from each (midpoint)? What is the electric field at this point?

Electric Potential (V)

v = = + k q r

k q1 r1

k q2 r2

v = (9.0x109Nm2/C2)(+50x10-6C) (10x10-2m)v = 4.5x106 Nm/Cv = 4.5x106 J/C v = 4.5x106 Volts

Electric Field (F)Electric Field (F)

ET = E Vectors

E1 =k q1 r2

= (9.0x109Nm2/C2)(50x10-6C)(10x10-2m)2

E1 = 4.5x107 N/C directed away from q1

Similarly E2 = 4.5x107 N/C directed away from q2

q1 q2

10cm10cm

E2

E1

ET = 0 N/C

Example ProblemExample Problem (Electric Potential : 2 Dimensions)

Calculate the electric potential of point ACalculate the electric potential of point A

v = k q

rv = k +( )q1

r1

q2

r2

v = (9.0x109Nm2/C2)( )

-50x10-6C + 50x10-6C 60x10-2m 30x10-2m

v = (9.0x109Nm2/C2)(-8.3x10-5C/m + 1.66x10-4C/m)v = (9.0x109Nm2/C2)(8.3x10-5C/m)v = 7.5x105Nm/Cv = 7.5x105J/C

v = 7.5x105 V

60cm30cm

52cm

A

q2 = +50 C q1 = -50 C

EAl

EA2

Equipotential Lines (Surfaces)Equipotential Lines (Surfaces)GravitationalGravitational

Line of Equipotential is merely a line (surface in 3-dimensional system) where potential energy remains constant as an object moves along the line

Lines of Equipotential are perpendicular to force field lines so that no work is done when the object moves

W = F d cos

Earth

Lines of Equipotential

Gravitational Field Lines

Equipotential Lines (Surfaces) Equipotential Lines (Surfaces) ElectrostaticElectrostatic

Equipotential Lines (surfaces)Equipotential Lines (surfaces)• encircle the charged particle• perpendicular to field lines• never cross each other because field lines never cross each other

Electric Field LinesElectric Field Lines• directed away from a positive charge (i.e. direction a positive “test charge” would move)• closer together indicates greater E-field• exit perpendicular to the surface• never cross each other

Test YourselfTest YourselfDraw E-fields and lines (surfaces) of Equipotential for the following situations.

c)

+q -q

a)

-q

b)

+q+q

Current and ResistanceCurrent and Resistance

(Electric) Current (I): The rate at which charge is flowing (through a wire).

1 Ampere = 1 Amp = 1 A = 1 Coulomb/sec

I = Qt

Conventional Current:Conventional Current: The flow of positive charge.

If conventional current is flowing to the right, then in reality, electrons are flowing to the left.

Positive charge, protons, don’t move.

Ampere (A): SI unit for current

Resistance:Resistance: A measure of what must be overcome to make charge flow

Ohm (Ohm ():): SI unit for resistance

1 ohm = 1 = 1 Volt/Amp

R = VI

If a large current results from a small V, then the resistance must be small.

If a small current results from a large V, then the resistance must be large.

Example ProblemExample ProblemA flow of 24 coulombs of charge passes through a wire in 2.1 seconds where a voltage of 37 volts is applied across the wire. Calculate a) how many charges moved through the wire, b) the current in the wire, and c) the resistance of the wire.Strategy:Strategy:Extract the data.Extract the data.

Q = 24 coulombst = 2.1 secondsV = 37 V

Apply Proper FormulasApply Proper Formulasa) 24 coulombs 1 electrons 1.60x10-19 coulombs

1.5x1020electrons

b) I = =Q 24 coulombs t 2.1 seconds

I = 11 coulombs/sec 11 amps

c) R = =V 37 Volts I 11 amps

3.4 ohmsR = 3.4 Volts/amp

ResistanceResistance

What factors might affect the resistance of a wire?

lA

R

R = lA

= resistivity of the material = ohms -meterResistivity is another physical property of a material

l

A

Resistors and Energy LossResistors and Energy LossIt is really friction in the wire that results in resistance to flow of charge. Friction causes heat.A resistor should get hot when voltage causes a current passes through it (Ex: filament in a light bulb)

Derivation:Derivation: Voltage = Joules/coulomb Current = Coulomb/sec

Voltage x Current

V x I = Joules/sec = WattsPower has units of Joules/sec or Watts

P = IV

P = I2R

P =?

V = IR

Joules x CoulombCoulomb sec=