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Physics 2B – Electric Forces and Electric Fields, Ch. 18

2-4 September 2014

•Assignment due 9/9/2014Due next week

Read Chapter 18F14 Problems 18.4, 18.10, 18.15, 18.21, 18.30, 18.37, 18.40, 18.47, 18.56,

18.59 (CJ 9th ed) Problem 18.6 means Problem 6 at the end of Chapter 18

Some figures and examples are from Serway and Vuille, College Physics, Brooks / Cole and Cutnell & Johnson, Physics, Wiley.

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Agenda• Electric Forces and Electric Fields

BackgroundCharged Objects and the Electric ForceConductors and InsulatorsCharging by Contact and InductionCoulomb’s LawElectric FieldElectric Field LinesGauss’ Law

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Background• Electric Charge e =1.60 x 10-19 coulomb (C)

Positive charge (+e) for protonsNegative charge (-e) for electrons

• Electrically Neutral: no net charge or number of protons equals number of electrons

• Electric charges are quantized.

Total charge q = Ne, N an integer

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Background (cont’d)• Law of Conservation of Electric Charge: net

electric charge of an isolated system remains constant.

Example: Ebonite gains electrons while animal fur loses electrons. Ebonite and animal fur are an isolated system.

• Electrically charged objects exert electrostatic forces upon each other.

Unlike charges attract each other (a) and like charges repel each other (b, c).

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Electrical Conduction / Insulation • Electric charges can move (conduct) through objects.

• Electrical conductor charges flow readily• Electrical insulator charges flow poorly• Why? Outermost electrons (valence electrons) can be detached and

result in charge flow.

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Charging Objects• Methods of Charging Objects:

1. Charging by Contact: charging an object by contacting another charged object

2. Charging by Induction: Giving an electric charge to another object without touching it

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Coulomb’s Law• Particles exert attractive or

repulsive forces upon each other.• Coulomb’s Law: the electrostatic

force between two charges q1 and q2 separated by a distance r is given by= | |

k = 8.99 x 109 N-m2 / C2

Note: k = 1 / (4π εo) where the permittivity of free space εo = 8.85 x 10-12 C2 / (N-m2 )

• The electrostatic force is directed along the line joining the charges.

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• Attractive force: q1 and q2have opposite signs

• Repulsive force: q1 and q2have same signs

If there are more than two charges, the net force is the vector sum of the individual forces.

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Electrostatic Force – Problem C18.14• The drawings show three charges that have the same magnitude but may have

different signs. In all cases the distance d between the charges is the same. The magnitude of the charges is |q| = 8.6 μC, and the distance between them is d = 3.8 mm. Determine the magnitude of the net force on charge 2 for each of the three drawings.

Equations:1. q = Ne, e =1.60 x 10-19 C2. = | |• k = 8.99 x 109 N-m2 / C2

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Electrostatic Force – Problem C18.22• An electrically neutral model airplane is flying in a horizontal circle on a 3.0-m

guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is 50.0 J. Reconsider the same situation, except that now there is a point charge of +q on the plane and a point charge of –q at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is 51.8 J. Find the magnitude of the charges.

Equations:1. q = Ne, e =1.60 x 10-19 C 2. = | |• k = 8.99 x 109 N-m2 / C2

Electric Field• Consider the net electrostatic force

experienced by a charge qo placed close to other charged objects.

• Electric Field E: the net electrostatic force F at a point experienced by a small test charge qo at that point divided by the charge itself.=

Electric field E is a vector.Direction of electric field is the same as the direction of the net force F acting on a positive test charge. (In other words, it points away from a positive charge).SI unit: Newton per Coulomb (N / C)

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Electric Field – Problem C18.43• A small object has a mass of 3.0 x 10-3 kg and a charge of -34 μC. It is placed at a

certain spot where there is an electric field. When released, the object experiences an acceleration of 2.5 x 103 m/s2 in the direction of the +x axis. Determine the magnitude and direction of the electric field.

Equations:1. q = Ne, e =1.60 x 10-19 C2. = | |• k = 8.99 x 109 N-m2 / C2

3. E = F / qo

Electric Field – Point Charge

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• Remember: the electric field E = F / |qo|.• However, Coulomb’s Law states that F = k |qo ||q| / r2

• Therefore, the electric field due to a point charge is

• E = k |q| / r2

• Consider two charges q1 = +16 μC and q2 = +4 μC. At what point P between the two charges is the net electric field zero?

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Electric Field – Parallel Plate Capacitor• Consider a parallel plate capacitor

consisting of two parallel metal plates, each with area A, separated by a vacuum. One metal plate has a net uniform charge of +q and the other plate, -q.

• We can show using Gauss’ Law (later) that the electric field is given by

E = q / (εo A) = σ / εoεo is the permittivity of free space = 8.85 x

10-12 C2 / (N-m2 )A is the area of the plateσ is the charge density q / A

• Note: the field is independent of the location. 13

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Electric Field – Problem C18.36• The membrane surrounding a living cell consists of an inner and

an outer wall that are separated by a small space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge density on the inner and outer walls has a magnitude of 7.1 x 10-6 C/m2 . (a) What is the magnitude of the electric field within the cell membrane? (b) Find the magnitude of the electric force that would be exerted on a potassium ion placed inside the membrane.

Equations:1. q = Ne, e =1.60 x 10-19 C2. = | |• k = 8.99 x 109 N-m2 / C2

3. E = F / qo4. E = kq / r2 point charge5. E = q / (εoA) parallel pl cp• εo = 8.85 x 10-12 C2 /(N-m2)

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Electric Field Lines• Electric Field Line: map of electric field (lines of force) in the

space surrounding electric charges.

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Positive point charge +qPositive test charge at positions 1-8

Negative point charge -qPositive test charge at positions 1-8

Electric Field Lines – CommentsElectric field lines start at positive charges and end at negative charges (point away from positive charges).Tangents to the field lines indicate direction of electrostatic forces.Electric field lines are closer to each other where the electric field is stronger.More electric field lines leave / enter a charge for larger charge magnitudes. Example: three times as many field lines leave a charge of +6q as would leave a charge of +2q.Electric field lines are in three-dimensions.Parallel electric field lines indicate that the electric field has the same direction and same magnitude at all points in space. Example: parallel plate capacitor.

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Electric Field Lines – Conceptual Understanding

Consider again the electric field lines for one positive and one negative point charge.What happens to the electric field lines if both charges are positive?What happens to the electric field lines if both charges are negative?

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There are three things wrong with part (a) of the drawing. What are they?

Conceptual Example – Drawing Electric Field Lines

•Start at positive charges and end at negative charges.•Tangents to the field lines indicate direction of electrostatic forces.•Closer to each other where the electric field is stronger.•More electric field lines leave / enter a charge for larger charge magnitudes.•Parallel electric field lines indicate that the electric field has the same direction and same magnitude at all points in space.

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Electric Field Lines – Problem C18.63• Charges of -4q are fixed to diagonally opposite corners of a square. A charge of

+5q is fixed to one of the remaining corners, and a charge of +3q is fixed to the last corner. Assuming that ten electric field lines emerge from the +5q charge, sketch the field lines in the vicinity of the four charges.

-4q

+3q -4q

+5q

•Start at positive charges and end at negative charges.•Tangents to the field lines indicate direction of electrostatic forces.•Closer to each other where the electric field is stronger.•More electric field lines leave / enter a charge for larger charge magnitudes.•Parallel electric field lines indicate that the electric field has the same direction and same magnitude at all points in space.

Electric Field Inside a Conductor

Consider a sphere composed of a conducting material such as copper (a). If there are excess negative charges (or positive) inside the sphere, the charges would experience repulsionand, therefore, travel to the surface of the sphere (b).Therefore, the inside of the sphere would be electrically neutral, and there would be no net movement of the negative charges inside the sphere.In other words, the electric field is zero at any point inside the conducting sphere.

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Gauss’ LawElectric fields are created by point charges.What is the effect of distributed or multiple point charges? Use Gauss’ Law.Gauss’s Law: relationship between a charge distribution and the resultant electric fieldThe product of the of the electric field E and the area A perpendicular to the field is the electric flux ΦE.

ΦE = Σ (E cos φ ) ΔA = Q / εo

Q is the net charge enclosed by a Gaussian surfaceGaussian surface: a closed surface (e.g., complete egg shell)A is the surface area of the Gaussian surfaceφ is the angle between the normal to the surface A and the electric field vector Eεo = 8.85 x 10-12 C2 / (N-m2 ) the permittivity of free space

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Application: Gauss’ Law for a Positive Point Charge and Spherical Gaussian Surface

Gauss’ Law: ΦE = Σ (E cos φ ) ΔA = Q / εo

Remember: E = kq / r2 = q / (4π εor2)However, for a sphere, A = 4π r2

Therefore, E = q / (A εo )

Or, EA = q / εo εo = 8.85 x 10-12 C2 / (N-m2 )

Or, for a sphere, electric flux ΦE = EA = q / εo

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Gauss’ Law – Problem C18.54• A spherical surface completely surrounds a collection of

charges. Find the electric flux through the surface if the collection consists of (a) a single +3.5 x 10-6 C charge, (b) a single -2.3 x 10-6 C charge, and (c) both of the charges in (a) and (b).

Equations:1. q = Ne, e =1.60 x 10-19 C2. = | |• k = 8.99 x 109 N-m2 / C2

3. E = F / qo4. E = kq / r2 point charge5. E = q / (εoA) parallel pl cap• εo = 8.85 x 10-12 C2 /(N-m2)6. ΦE = Σ (E cos φ ) ΔA = Q / εo

Gauss’ Law for a Parallel Plate Capacitor

Remember, for a charge distribution and arbitrary Gaussian surface,

Electric flux ΦE = Σ (E cos φ ) ΔA = Q / εo

Example: Parallel Plate CapacitorChoose the Gaussian surface to be a closed cylinder that starts within the left plate and ends between the plates.Then, ΦE = ΦE1+ ΦE2+ ΦE3

Or, ΦE = 0 (no field within plate) + 0 (since φ = 90o) + Σ(E cos φ) ΔAOr, ΦE = Σ(E cos 0) ΔA = E*1*A = EAOr, EA = Q / εo E = (Q / A) / εo

Or, E = σ / εo (as stated earlier on slide 13)24

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Gauss’ Law – Problem C18.58• A charge Q is located inside a rectangular box. The electric

flux through each of the six surfaces of the box is: Φ1 = +1500 N-m2/C, Φ2 = +2200 N-m2/C, Φ3 = +4600 N-m2/C, Φ4 = -1800 N-m2/C, Φ5 = -3500 N-m2/C, and Φ6 = -5400 N-m2/C. What is Q?

Equations:1. q = Ne, e =1.60 x 10-19 C2. = | |3. k = 8.99 x 109 N-m2 / C2

4. E = F / qo5. E = kq / r2

6. E = q / (εoA)• εo = 8.85 x 10-12 C2 /(N-m2)7. ΦE = Σ (E cos φ ) ΔA = Q / εo

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Summary• Electric Forces and Electric Fields

Background q = NeCharged Objects and the Electric ForceConductors and InsulatorsCharging by Contact and Induction

Coulomb’s Law = | |Electric Fields = = k |q| / r2 point charge, E = q / (εo A) = σ / εo parallel plate capacitor Electric Field LinesGauss’ Law ΦE = Σ (E cos φ ) ΔA = Q / εo