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Study Guide 8: Electricity Text: chapter 15, sections 2-3 and 5. (Note, the order in which we cover this material differs from the Study Guide, see Course Outline. We start with section 15.5) Quiz 5 (final day, Friday, March 28, 2008) you should know: Material from Computer lab exp. 7 (Radiation) and exp. 8 (Electricity): in particular From semilog and log-log graphs of provided data (exp. 7); 1. How to calculate the linear attenuation coefficient. 2. Determine half-time T1/2. 3. Study the exponent in power-law decay from a point radioactive source. 4. Apparatus used in exp. 8 (bottom of page 8-1 of lab manual) and how to arrange this to study parallel and series circuits. 5. Using 4 to calculate current and resistances.

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ELECTRIC CURRENT

The 3 quantities describing a DC circuit are:

1) Current (I) 2) Electric Potential (V)

3) Resistance(R)

Definitions: Current=flow of charge, Δq, through a cross-sectional area in time Δt.

SI unit = Ampere (A) = (1 Coulomb/second). Motion of electrons (negative!) actually produces the current. E.g., in a copper wire:

I = Δq/Δt

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Copper atomic cores (ignore their motion)

Direction of Conventional Current

Free Electrons A negative current or a positive current moving in the opposite direction are equal mathematically. Therefore, we treat all currents as the positive flow of charge in problem solving. Electric Potential - symbol VAB If the work in moving a charge q from a point B to a point A is WAB, then

where VAB = VA−VB SI unit = Joules/Coulomb = Volt (V). Electric potential is often called the voltage.

VAB = WAB/q

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Resistance = applied voltage required per current - depends on material. Consider a voltage, VAB, across a material with resistance R: R A B I

VAB = VA − VB If R is a constant, the above quantities obey

. Ohm's Law UNITS FOR R: Volt/Amperes = Ohms (Ω). (Conductance (unit: Siemens, S = Ω−1) is the inverse of resistance,

g = 1/R )

VAB = IR

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DC CIRCUITS: • Current varies sinusoidally on power grids - Alternating

Current (AC) • We study constant current - Direct Current (DC).

R

A simple DC circuit I

+ − V (note the + terminal of battery is

at higher potential than the −)

Kirchoff's rules

1. The current entering a point is equal to the current leaving it.

2. The net change in voltage around a closed loop is

zero. Other useful observations: • points connected by a PERFECT resistor are at the

same electric potential.

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• Resistors arranged in parallel and series have specific

rules to determine their effective resistance. PARALLEL NETWORK: R1 I1 R2 I I A I2 R3 I3

V The equivalent resistance, Req, such that V = IReq, for the 3 resistors above is found using Kirchoff's first rule at point A,

Current in I = I1 + I2 + I3

Current out

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NB All 3 resistors have same voltage across them, V, (parallel). Therefore V = I1R1 V = I2R2 ∴ I = V/R1 + V/R2 + V/R3 V = I3R3 = (1/R1 + 1/R2 + 1/R3)V so that Req = (1/R1 + 1/R2 + 1/R3)-1 SERIES NETWORK: I R1 R2 R3 A B V1 V2 V3

VAB = VA−VB

Ohm’s law: IReq = VAB. Notice…

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1. Each resistor now experiences a different V. 2. VAB = V1+V2+V3 3. Current out = current in, so each resistor has the

same current I. Ohm’s law for each resistor;

IR1 = V1

IR2 = V2

IR3 = V3

∴ IReq = IR1 + IR2 + IR3 . Dividing by I on both sides, we find the equivalent resistance,

Req = R1 + R2 + R3

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Sample Problem 1: Parallel Resistors Calculate Req for the following network of resistors: R1 = 50.0 Ω R2 = 33.3 Ω R3 = 16.6 Ω As shown earlier, Req = (1/R1 + 1/R2 + 1/R3)-1. Therefore, Req = (1/50.0 + 1/33.3 + 1/16.6)−1 = (0.02 + 0.03 + 0.06)−1. That is, Req = 9.1 Ω.

Sample Problem 2: Resistors in Series Calculate Req for the same three resistors in a series arrangement: R1 R2 R3 Ans. Req = 99.9 Ω.

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Sample Problem 3: Series/Parallel Resistors Calculate Req for the following combination of resistors: R1 = 50.0 Ω R3 = 100.0 Ω R2 = 25.0 Ω R4 = 166.7 Ω

Ans. Req = 79.2 Ω.

Sample Problem 4: Series, Parallel or What?

Calculate Req for the following network. All the resistors are 2.0 Ω, while the wires are assumed to be perfect resistors.

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This is a parallel-series arrangement. Using methods similar to example 3 to solve for Req, we find that Req = 5/3 = 1.67 Ω. CALCULATING THE RESISTANCE: Resistance depends on the dimensions and material through which the current is flowing. For example, consider a conductor with cross-sectional area A and length L; L A Then, the resistance is given by,

ρLRA

=

where ρ is the resistivity of the material and can be found in tables of scientific data. For metals ρ ≈ 10-8 Ω⋅m; for insulators ρ ≈ 10+13 Ω⋅m (more than 10+20 times that of a metal!).

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As hinted earlier, resistivity depends upon the collisions of the current carriers with the ionic cores making up the bulk of the material. Clearly this will depend on the temperature of the material. The higher the temperature, the greater the ionic core oscillations. These oscillations impede the flow of electrons contributing to the resistance, so usually the resistivity increases with temperature. Sample Problem 5: Resistance of a cylinder Calculate the resistance of the cylinder below (be careful about the units !):

2.00 cm

(diameter) 0.500 mm ρ = 1.50 × 10-2 Ω⋅cm

Ans. R = 15.3 Ω

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For additional problems involving circuits with batteries, recall Kirchoff’s rules: 1. The current entering a point is equal to the current

leaving it. (The “point rule”) 2. The net change in voltage around a closed loop is

zero. (The “loop rule”) These rules must be supplemented by important sign conventions (see text page 15-19).

Example 3 V 1 Ω 20 Ω e f I2 I2 loop 1 I2 c d 5 V I1 1 Ω I1 I3 loop 2 I3 a b 5 Ω I3

(a) What is the current, I1, through the middle

branch? (b) What is Vab = (Va - Vb)?

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GENERAL STRATEGY – MESH EQUATIONS: Use Kirchoff's rules and Ohm's Law to solve for the unknown loop currents, I1, I2, and I3, then determine Vab. (1) Use Kirchoff's Point Rule (#1) to relate currents.

E.g., at point c:

I1 = I2 + I3 or I3 = I1 - I2

(2) Use Kirchoff's Loop Rule (#2) on as many loops

as required to generate sufficient equations. Two loops are required from above.

(i) For loop 1: going clockwise starting at c (but this choice of direction is arbitrary)

-3V - I2 (1 Ω) - I2 (20 Ω) - I1 (1 Ω) + 5V = 0 “-” because going from + to – terminal of the battery The IR terms are negative since each resistor is traversed in the same direction as the current (this is one of the sign conventions).

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∴ - I1 - 21 I2 = -2 Amps ∴ I1 + 21 I2 = 2 A [1]

(ii) For loop 2: going clockwise starting at c (and using I3 = I1−I2)

- 5 V + I1(1 Ω) + (I1 - I2)(5 Ω) = 0 Now the IR terms are positive because each resistor is traversed in the opposite direction to the current. ∴ 6 I1 - 5 I2 = 5 A [2]

(3) Solve the resulting system of equations:

Multiplying equation [1] by 5:

5 I1 + (5)(21)I2 = 10 A [3]

Multiplying equation [2] by 21: (21)(6) I1 − (5)(21)I2 = 105 A [4]

Add [3] and [4] (terms in I2 cancel) …

131 I1= 115 A

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∴ I1 = 115/131 = 0.878 A (ans. To (a))

We’ll also need I2: from [2], I2 = (6I1 − 5)/5 = (6)0.878/5 -1 = 0.0536 A For part (b), Vab = (I1 − I2) (5 Ω) = 4.12 V. This is also equal to Vcd and Vef because these voltages are all parallel. (Check: Vcd = 5V − I1(1 Ω ) = 5V − 0.878V = 4.12 V (The IR term is negative because I1 is toward the left.) Vef = 3V + I2(1Ω +20Ω) = 3V + 0.0536(21) = 4.126 V)

ELECTRIC FORCE AND FIELD

The underlying concepts from the last section are somewhat explained. E.g., Ohm's law stems from the electric forces between two charges (Coulomb's law) General Information: • The SI unit of charge is the Coulomb (C). • The smallest magnitude of charge that can occur

outside of a nucleon is that on a single electron, = 1.602 × 10 -19 C.

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• Charge is either positive or negative. Charges of the same sign (-,- or +,+) repel and charges of opposite sign (+,- or -,+) attract.

ELECTRIC FORCE-COULOMB'S LAW: F21 q1 F12

q2 NB: Equal attracting force on EACH charge! (These forces are in opposite directions due to Newton’s 3rd law)

• direction of 12F is given by the signs of the charges.

• magnitude is given by Coulomb's law.

• If there are more charges, add forces together as VECTORS. (But, although there are examples in the text and study guide where the charges are arranged in a non-linear manner (e.g., example 15-1 in text and on page SG 8-5), in practice we only deal with linear arrangements of charges, which involves only algebraic addition of forces.

Assuming nothing (not even air) lies between the charges:

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( )= 1 2

12 212

k q qF

r

where k = 9.0 × 109 Nm2C-2 is the constant for the vacuum. If the charges are not in a vacuum then Coulomb's law takes the form:

( )1 2

12 2

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k qqF

rK=

where K is the (unitless) dielectric constant for the medium in which the charges are immersed. For example:

water K = 80.4 air K = 1.0005 wood K = 5.0

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ELECTRIC FIELDS: Each charge creates an electric field around itself. The interaction of a second charge with this electric field gives rise to the electric force. For example,

Electric field lines:

The magnitude of the electric field at distance r from a point charge q is given by:

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k qE =

rK

where k and K have the same meanings as in Coulomb’s Law. The direction of the field = direction of the force on small positive charge at that point.

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If there are multiple sources, the total field = the VECTORIAL sum of the electric fields from all sources.

FORCE ON A CHARGE q1 IN AN ELECTRIC FIELD E:

1F qE=.

NB: electric field and force point in the same direction for a positive charge q1 and in the opposite direction for a negative charge.

ELECTRIC POTENTIAL ENERGY and POTENTIAL:

Consider a small positive charge q1 in a constant electric field E. It will experience a force, q1E, in the direction of E. To move it in the opposite direction, a force with magnitude q1E must be applied in that direction. To move the charge a distance d in that direction, work must be done against the electric force. Work(W) = force × distance for constant forces. 1W=Fd=qEd∴

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This energy depends on the charge q1. The electric potential, V, is defined to be independent of that charge, i.e.,

V = W/q1 or

W = q1V ( = electric potential energy U)

Potential due to a point charge The field due to a point charge isn't constant, and to get the work required to place two charges a distance r apart you must integrate the force over the path traveled. A point of reference must be chosen where V = 0. For point charges r = ∞ is assumed to have V = 0. The electric potential around a point charge q surrounded by a medium with dielectric constant K then takes the form,

k qV

rK=

The energy U (or work) required to move q1 from infinity to a distance r from q is

U = q1V

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Using the previous equation for the potential around q, we get

k q q1UrK

=.

Finally, the potential at a position due to many point charges qi is the algebraic sum of the potentials from each charge, i.e., i

iVV = ∑

Sample Problem 6 A sodium ion (singly charged positive ion) is 75 nm (1nm = 10-9m) from a chloride ion (singly charged negative ion) in the protoplasm of a cell (assume the protoplasm has the same dielectric constant as water). What forces do the ions exert on each other? Note that the sodium ion has a positive charge, qNa+

= 1.602 × 10 –19 C and the chloride ion has a negative charge, QCl

− = -1.602 × 10 –19 C.

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Ans. F = 5.11 × 10−16 N

This force is attractive (since the charges are of opposite sign), hence the forces on each particle have the directions shown on page 17. Sample Problem 7 Three ions (Ca2+, Na+ and Cl−) in a vacuum (hence K = 1) are in a linear arrangement shown below: Ca2+ Na+ Cl−

Find (a) The net electric field E at the position of the Na+

ion. (b) The net force on the Na+ ion. (c) The electric potential V at the position of the Na+

ion. (d) Repeat (a)−(c) at the position of the Cl− ion.

r23 =10 nm r12 =20 nm

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Answers: (a) E = 2.16×107 N/C (acting toward the right) (b) F = 3.46×10−12 N (acting toward the right) (c) V = 0. (d) (Without numbers)

F = qCl

−E (toward left)

( )

2+ +Ca Na

2312 23

kq kqV = +

rr +r

( )2+ +

2+ +Ca Na

2 2Ca Na2312 23

kq kqE= E + E = + (toward right)

rr +r