AP-C Electric Force and Electric Field

AP-C Objectives (from College Board Learning Objectives for AP Physics) 1. Charge and Coulomb’s Law a. Describe the type of charge and the attraction and repulsion of charges b. Describe polarization and induced charges. c. Calculate the magnitude and direction of the force on a positive or negative charge due to other specified point charges. d. Analyze the motion of a particle of specified charge and mass under the influence of an electrostatic force. e. Describe the process of charging by induction. f. Explain why a neutral conductor is attracted to a charged object. 2. Electric Field due to Point Charges a. Define the electric field in terms of force on a test charge. b. Describe and calculate the electric field produced by one or more point charges. c. Calculate the magnitude and direction of the force on a positive or negative charge placed in a specified field. d. Interpret electric field diagrams. 3. Gauss’s Law a. Understand the relationship between electric field and electric flux. i. Calculate the flux of an electric field through an arbitrary surface or of a uniform field over and perpendicular to a Gaussian surface. ii. Calculate the flux of the electric field through a rectangle when the field is perpendicular to the rectangle and a function of one coordinate only. iii. State and apply the relationship between flux and lines of force. b. Understand Gauss’s Law i. State Gauss’s Law in integral form and apply it qualitatively to relate flux and electric charge for a specified surface. ii. Apply Gauss’s Law, along with symmetry arguments, to determine the electric field for a planar, spherical, or cylindrically symmetric charge distribution. iii. Apply Gauss’s Law to determine the charge density or total charge on a surface in terms of the electric field near the surface. 4. Electric Fields due to Other Charge Distributions a. Calculate the electric field of a straight, uniformly charged wire; the axis of a thin ring of charge; and the center of a circular arc of charge. b. Identify situations in which the direction of the electric field produced by a charge distribution can be deduced from symmetry considerations. c. Describe the patterns and variation with distance of the electric field of oppositely charged parallel plates; a long uniformly charged wire; a thin cylindrical shell; and a thin spherical shell. d. Determine the fields of parallel charged planes, coaxial cylinders, and concentric spheres.

- 1 - Charge and Coulomb's Law

Electric Charge AP-C Objectives (from College Board Learning Objectives for AP Physics) Electric charge (q) is a fundamental property of certain particles. The smallest 1. Charge and Coulomb’s Law amount of isolatable charge is the elementary charge (e), equal to 1.6×10-19 a. Describe the type of charge and the attraction and repulsion of charges coulombs. Charge can be positive or negative. b. Describe polarization and induced charges. c. Calculate the magnitude and direction of the force on a positive or negative charge due to other specified point charges. d. Analyze the motion of a particle of specified charge and mass under the influence of an electrostatic force. Atomic Particles Conductors and Insulators e. Explain the process of charging by induction. protons have a charge of +1e Charges can move freely in conductors. electrons have a charge of −1e Charges cannot move freely in insulators. f. Explain why a neutral conductor is attracted to a charged object. neutrons are neutral 2. Electric Field due to Point Charges Atoms with an excess of protons a. Define the electric field in terms of force on a test charge. or electrons are known as ions. b. Describe and calculate the electric field produced by one or more point charges. c. Calculate the magnitude and direction of the force on a positive or negative charge placed in a specified field. Polarization and Electric Dipole Moment d. Interpret electric field diagrams. When a charged object is brought near a conductor, the electrons in the conductor are free to move. When a charged object is brought near an insulator, the electrons are not free to move, but they may spend a little more time on one side of their orbit than another, creating a net separation of charge in a process known as Electric Field (E) Electric Force and Coulomb’s Law polarization. The distance between the shifted positive and negative charges, The electric field describes the amount of electrostatic multiplied by the charge, is known as the electric dipole moment. Like charges repel, opposite charges attract. force observed by a charge placed at a point in the conductor insulator field per unit charge. The electric field vector points in 2 1 q q −12 - 1 2 C + - + - + the direction a positive test charge would feel a force. F = ε = 8.85×10 2 + 2 0 + - + - + Electric field strength is measured in N/C, which are N i m + + 4πε0 r + - + + - + equivalent to V/m. + - + + - + - + - + F 1 q + + + - + - + E F qE E charged rod - charged+ rod = = = 2 + + + - + + + q 4πε0 r + + + + Electric field lines indicate the direction of the electric force on a positive test charge.

Conduction and Induction Charging by contact is known as conduction. If a charged conductor is brought into contact with an identical neutral conductor, the net charge will be shared across the two conductors. Charging an object without placing it in contact with another charged object is known as induction.

Electric Field due to Multiple Point Charges The electric field follows the law of superposition. In order to determine the electric field due to multiple point charges, add up the electric field due to each of the individual charges.

Sample Problem: E Field due to Point Charges

Find the electric field at the origin due to the three point charges shown.

y (m) 1 q 1 (2) 8 E1 = 2 = 2 =< 0,−2.81×10 > 9 4πε0 r 4πε0 8 8 +2C 1 q 1 (1) 9 N E = = = 1.13×10 C → 7 2 2 2 2 2 4πε0 r 4πε0 ( 2 + 2 ) E E E E 6 9 N 9 N tot = 1 + 2 + 3 → E2 =< −1.13×10 C × cos45°,−1.13×10 C × sin45° >→ 5 8 9 8 8 N E =< −7.95×10 ,−7.95×10 > N Etot =< −5.14 ×10 ,−1.08 ×10 > C 4 2 C 1 q 1 2 3 E 2.81 108 ,0 N +1C 3 = 2 = 2 =< × > C 2 4πε0 r 4πε0 8 1 -2C x (m)

1 2 3 4 5 6 7 8 9

Sample Problem: Where is the E Field Zero?

Determine the x-coordinate where the electric field is zero using the diagram. 1 q E = 1 1 4πε r 2 r 11-r 0 +1C +2C x (m) 1 q2 E = 2 4 (11 r)2 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 πε0 −

1 q k= 1 1 1 q2 4πε0 k(1) k(2) 2 Etot = E1 + E2 = 2 − 2 ⎯⎯⎯→ 2 − 2 = 0 → r + 22r −121= 0 → r = 4.56 4πε0 r 4πε0 (11− r) r (11− r) x = −6 + 4.56 = −1.44

- 2 - Electric Fields due to Other Charge Distributions 1

Charge Densities AP-C Objectives (from College Board Learning Objectives for AP Physics) ΔQ 1. Electric Fields due to Continuous Charge Distributions Linear Charge Density λ = a. Calculate the electric field of a straight, uniformly charged wire; the axis of a thin ring of ΔL charge; and the center of a circular arc of charge. b. Identify situations in which the direction of the electric field produced by a charge distribution ΔQ can be deduced from symmetry considerations. Surface Charge Density σ = ΔA ΔQ Volume Charge Density ρ = ΔV

E-Field Due to a Thin Uniform Semicircle of Charge A thin insulating semicircle of charge Q with radius R is centered around point C. Determine the electric field at point C due to the semicircle of charge.

Symmetry Arguments Horizontal component will cancel out since the charge is uniformly distributed, so we only need to worry about the vertical component of the electric field.

 1 dQ  θ=π 1 λRdθ ΔQ Q dE = sinθ ⎯d⎯Q=λ⎯Rdθ⎯→ dE = sinθ dθ → λ = = y 2 ∫ y ∫θ=0 2 ΔL πR 4πε0 R 4πε0 R  λ θ=π  λ π 2λ λ E = sinθ dθ → E = (−cosθ) = = y ∫θ=0 y 0 4πε0 R 4πε0 R 4πε0 R 2πε0 R

E Field Due to a Thin Straight Insulating Wire Find the electric field some distance d from a long straight insulating rod of length L at Q λ = a point P which is perpendicular to the wire and equidistant from each end of the wire. L

Strategy 1. Divide the total charge Q into smaller charges ΔQ. 2. Find the electric field due to each ΔQ. 3. Find the total electric field by adding up the individual electric fields due to each ΔQ. 4. Realize the y-component of the electric field is 0 due to symmetry arguments.

2 2 1 ΔQ ri = yi +d Ei = Ei cosθi = 2 cosθi ⎯⎯⎯d ⎯d⎯⎯→ x cosθ = = 4πε r i 2 2 0 i ri yi +d

Q Q dy 1 ΔQ d 1 dΔQ Δ = L Ei = 2 2 = 3 ⎯⎯⎯⎯→ x 4πε ( y + d ) 2 2 4πε 2 2 2 0 i yi + d 0 ( yi + d )

y L 1 dQdy = 2 dQ / L dy E E i = 3 → x = L 3 → x 2 2 y=− 2 2 4πε 2 ∫ 2 4πε 2 0 L( yi + d ) 0 ( y + d ) dx x = +C L ∫ 3 2 2 2 dQ / L 2 dy 2 2 2 a a +x E (a +x ) x = L 3 ⎯⎯⎯⎯⎯⎯⎯⎯→ ∫− 2 2 2 4πε0 2 ( y + d )

L ⎛ 2 ⎞ dQ / L y Q / L ⎡ L / 2 −L / 2 ⎤ Ex = ⎜ 1 ⎟ = ⎢ 1 − 1 ⎥ → 4πε 2 2 2 2 4πε d L 2 2 2 − L 2 2 2 0 ⎜ d ( y + d ) L ⎟ 0 ⎢(( 2 ) + d ) (( 2 ) + d ) ⎥ ⎝ − 2 ⎠ ⎣ ⎦ Q / L L Q Ex = 1 = 4πε d L 2 2 2 L 2 2 0 (( 2 ) + d ) 4πε0d ( 2 ) + d

What is the E Field if the rod is infinite?

Q ⎛ ⎞ λ= Q Q Q L λ Ex = lim⎜ ⎟ = = ⎯⎯⎯→ Ex = L→∞ ⎜ L 2 2 ⎟ 4πε d( L ) 2πε dL 2πε d ⎝ 4πε0d ( 2 ) + d ⎠ 0 2 0 0

Note that Q approaches infinity as L approaches infinity!

What is the E field if the distance d is infinite? ⎛ ⎞ Q Q ⎜ ⎟ Acts like the E-field of a point charge! Ex = lim = 2 d→∞ ⎜ 2 2 ⎟ 4 d 4 d L d πε0 ⎝ πε0 ( 2 ) + ⎠

- 3 - Electric Fields due to Other Charge Distributions 2

Charge Densities AP-C Objectives (from College Board Learning Objectives for AP Physics) ΔQ 1. Electric Fields due to Continuous Charge Distributions Linear Charge Density λ = a. Calculate the electric field of a straight, uniformly charged wire; the axis of a thin ring of ΔL charge; and the center of a circular arc of charge. b. Identify situations in which the direction of the electric field produced by a charge ΔQ Surface Charge Density σ = distribution can be deduced from symmetry considerations. ΔA ΔQ Volume Charge Density ρ = ΔV

Electric Field on the Axis of a Thin Ring of Charge Q Q λ = = ΔQ = λRdφ Find the electric field at a point on the axis (perpendicular to the ring) of a thin L 2πR insulating ring of radius R that is uniformly charged as shown in the diagram.

Symmetry Arguments

By symmetry, the only net electric field will be in the z-direction.

 2 2 1 ΔQ ri = z +R 1 ΔQ z Ei = Ei cosθi = 2 cosθi ⎯⎯⎯z ⎯z⎯⎯→ Ei = 2 2 1 → Z cosθ = = Z 2 2 2 4πε0 r i r 2 2 4πε0 (z + R ) (z R ) i i z +R +

φ=2π φ=2π 1 z ΔQ=λRdφ 1 z 1 z Ei = 3 ΔQ ⎯⎯⎯⎯→ Ez = 3 λR dφ = 3 λR dφ → Z 2 2 2 2 2 2 ∫φ=0 2 2 2 ∫φ=0 4πε0 (z + R ) 4πε0 (z + R ) 4πε0 (z + R ) Q λ= 1 z 2πR 1 zQ Ez = 3 λR(2π) ⎯⎯⎯→ Ez = 3 2 2 2 2 2 2 4πε0 (z + R ) 4πε0 (z + R )

Electric Field Due to a Uniformly Charged Disk Find the electric field due to a uniformly charged insulating disk of radius R at a point P perpendicular to the disk as shown in the diagram. Q Q σ = = A πR2 Symmetry Arguments

By symmetry, the only net electric field will be in the z-direction.

ΔQ = σΔA = σ2πrΔr zΔQ z 2 r dr i 1 i ΔQ=σ2πridr 1 σ π i Ei = 3 ⎯⎯⎯⎯→ Ez = 3 → Z 4πε 2 2 2 ∫ 4πε 2 2 2 0 (z + ri ) 0 (z + ri ) What is the E field if the disc is z2 +R2 2 2 infinite (an infinite plane)? σz R rdr 2 2 σz 1 z +R du σz ⎛ −2⎞ E u=z +r E z = 3 ⎯d⎯u=2⎯rdr→ z = 2 3 = 1 → ∫r=0 2 2 2 ∫u=z 2 ⎜ 2 ⎟ ⎛ ⎞ 2ε0 (z + r ) 2ε0 2 u 4ε0 ⎝ u ⎠ 2 σ z σ i z lim 1− = R→∞ 2ε ⎜ 2 2 ⎟ 2ε σz ⎛ −2 −2⎞ σz ⎛ 1 1 ⎞ σ ⎛ z ⎞ 0 ⎝ z + R ⎠ 0 E = − = − → E = 1− z ⎜ 2 2 ⎟ ⎜ 2 2 ⎟ z ⎜ 2 2 ⎟ 4ε0 ⎝ z + R z ⎠ 2ε0 ⎝ z z + R ⎠ 2ε0 ⎝ z + R ⎠

Electric Field due to a Finite Charged Rod Find the electric field due to a finite uniformly charged rod of length L lying on its side at some distance d away from the end of the rod.

Symmetry Arguments L d P Ex By symmetry, the only electric field will be in the x-direction. Q Q x λ = → ΔQ = λΔx → dQ = λdx L d+L x=d+L d+L 1 ΔQ 1 dQ dQ=λdx 1 λdx λ dx λ ⎛ −1⎞ Ei = 2 → Ex = 2 ⎯⎯⎯→ Ex = 2 = 2 → Ex = → x 4πε x ∫ 4πε x ∫x=d 4πε x 4πε ∫d x 4πε ⎜ x ⎟ 0 0 0 0 0 ⎝ ⎠ d λ ⎛ −1 −1⎞ λ ⎛ −d + d + L⎞ λ ⎛ L ⎞ Q L 1 Q E λ=Q/L E E x = ⎜ − ⎟ = ⎜ ⎟ = ⎜ ⎟ ⎯⎯⎯→ x = → x = 4πε0 ⎝ d + L d ⎠ 4πε0 ⎝ d(d + L) ⎠ 4πε0 ⎝ d(d + L)⎠ 4πε0 L d(d + L) 4πε0 d(d + L)

- 4 - Gauss's Law

AP-C Objectives (from College Board Learning Objectives for AP Physics) 1. Gauss’s Law a. Understand the relationship between electric field and electric flux. i. Calculate the flux of an electric field through an arbitrary surface or of a uniform field over and perpendicular to a Gaussian surface. ii. Calculate the flux of the electric field through a rectangle when the field is perpendicular to the rectangle and a function of one coordinate only. iii. State and apply the relationship between flux and lines of force. b. Understand Gauss’s Law i. State Gauss’s Law in integral form and apply it qualitatively to relate flux and electric charge for a specified surface. ii. Apply Gauss’s Law, along with symmetry arguments, to determine the electric field for a planar, spherical, or cylindrically symmetric charge distribution. iii. Apply Gauss’s Law to determine the charge density or total charge on a surface in terms of the electric field near the surface.

Electric Flux Through Open Surfaces Electric Flux Through Closed Surfaces Convention: Normals to closed surfaces point from the inside to the outside. Electric Flux (Φ) is the amount of electric field penetrating a surface. Total flux through the closed surface is positive if there is more flux from     dA inside to outside than outside to inside, and negative if there is more flux dΦ = E • dA = EdAcosθ → Φ = dΦ = E • dA E from outside to inside than inside to outside. dA ∫ ∫A   Φ = ∫ dΦ =∫ E • dA dA

Integral over closed surface

Derivation of Gauss’s Law

Consider a point charge inside a spherical shell of radius R. Determine the flux through the sphere. Gauss’s Law   Useful for finding the electric field due dΦ = E • dA = EdAcosθ ⎯⎯θ=0⎯→ dΦ = EdA → to charge distributions for cases of: cosθ=1 1) Spherical symmetry 2 Φ dΦ E dA E dA EA A=4πR 2) Cylindrical symmetry = ∫ =∫ = ∫ = ⎯⎯⎯→ E 3) Planar symmetry Q E= rˆ   Q 2 dA enclosed 2 4πε R 2 ⎛ Q ⎞ Q +Q Φ = E • dA = Φ = 4πR E ⎯⎯⎯0 ⎯→Φ = 4πR = → ∫ ⎜ 2 ⎟ ε0 ⎝ 4πε0 R ⎠ ε0   Q R Φ = ∫ E • dA = enclosed Gauss’s Law! ε0

Sample Problem: Electric Field due to a Thin Hollow Shell r Consider a thin hollow shell of uniformly distributed charge Q. Find the electric field inside and outside the sphere. o Q

Choose a “Gaussian Surface” as a sphere (first inside the shell of charge, then outside the shell of charge). ri By symmetry, the electric field at all points on the Gaussian spheres must be the same, and it must point radially in or out.

R Inside the shell of charge (ri

  Q Q Q Q 0 E • dA = enclosed → 4πR2 E = enc → E = enc ⎯⎯enc =⎯→ E = 0 E ∫ 2 ε0 ε0 4πε0 R

1 Outside the shell of charge (ro>R): r 2  Same answer as if all the  Qenclosed 2 Qenc Qenc charge Q was placed at a point E • dA = → 4πR E = → E = 2 ∫ in the center of the sphere. ε0 ε0 4πε0 R r R

Sample Problem: Electric Field due to an Infinite Plane Consider an infinite plane of uniform charge density σ. Determine the electric field due to the plane. A E Choose a “Gaussian Surface” as a cylinder as shown in the diagram. By symmetry, the electric field at all points on the cylinder must point perpendicular to the plane through the caps of the cylinder. d Q 2   Q σ= 0 enc A σA symmetry E • dA = ⎯Q⎯=σA⎯→ΦE + ΦE + ΦE = ⎯Φ⎯⎯=0⎯→ ∫ ε top bottom sides ε Esides Q Note: There is NO dependence 0 0 on distance from the plane! Φ =EA A σA Etop σA σ ΦE + ΦE = ⎯Φ⎯⎯=⎯EA→2EA = → E = top bottom Ebottom d ε0 ε0 2ε0

Sample Problem: Electric Field Between Parallel Charged Planes TOP BOTTOM NET Find the electric field surrounding and in between two oppositely-charged parallel planes or plates E E E 0 2 2 Strategy: Use E Field from a single infinite plane with surface charge density σ to 0 0 derive solution by adding the electric fields from each of the planes using the top superposition principle. E E E 2 2 Note that this is not accurate near the ends of the planes or plates. 0 0 0 bottom

E E E 0 2 0 2 0

Sample Problem: Electric Field due to Infinite Line of Charge Determine the electric field at a distance R from an infinitely long line of uniform charge density λ. R

Choose a “Gaussian Surface” as a cylinder centered around the line of charge as shown. By symmetry, the electric field at all points on the cylinder must point radially in or out.   Qenc Qenc symmetry Qenc ΦCyl =2πRLE E • dA = → ΦL + ΦR + ΦCyl = ⎯Φ⎯Φ⎯0⎯→ΦCyl = ⎯⎯⎯⎯→ ∫ L + R = ε0 ε0 ε0 Q Q λ= λL λ enc L Same answer as that reached using Coulomb’s Law 2πRLE = ⎯Q⎯=λL⎯→2πRLE = → E = ε0 ε0 2πε0 R

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