Lecture 20R: MON 13 OCT

Exam 2: Review Session CH 23.1–9, 24.1–12, 25.1–6,8 26.1–9

Physics2113 Jonathan Dowling

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Exam 2 •  (Ch23) Gauss’s Law

•  (Ch24) Sec.11 (Electric Potential Energy of a

System of Point Charges); Sec.12 (Potential of

Charged Isolated Conductor)

•  (Ch 25) Capacitors: capacitance and capacitors;

caps in parallel and in series, dielectrics; energy,

field and potential in capacitors.

•  (Ch 26) Current and Resistance: current, current

density and drift velocity; resistance and

resistivity; Ohm’s law.

•  Gauss’ law: Φ=q/ε0 . Given the field, what is the charge enclosed? Given the charges, what is the flux? Use it to deduce formulas for electric field.

•  Electric potential: –  What is the potential produced by a system of charges?

(Several point charges, or a continuous distribution) •  Electric field lines, equipotential surfaces: lines go from

+ve to –ve charges; lines are perpendicular to equipotentials; lines (and equipotentials) never cross each other…

•  Electric potential, work and potential energy: work to bring a charge somewhere is W = –qV (signs!). Potential energy of a system = negative work done to build it.

•  Conductors: field and potential inside conductors, and on the surface.

•  Shell theorem: systems with spherical symmetry can be thought of as a single point charge (but how much charge?)

•  Symmetry, and “infinite” systems.

At each point on the surface of the cube shown in Fig. 24-26, the electric field is in the z direction. The length of each edge of the cube is 2.3 m. On the top surface of the cube E = -38 k N/C, and on the bottom face of the cube E = +11 k N/C. Determine the net charge contained within the cube.���[-2.29e-09] C

Gauss’ law

Gauss’s Law: Cylinder, Plane, Sphere

Problem: Gauss’ Law to Find E

Gauss’ law

A long, non conducting, solid cylinder of radius 4.1 cm has a nonuniform volume charge density that is a function of the radial distance r from the axis of the cylinder, as given by r = Ar2, with A = 2.3 µC/m5.

(a) What is the magnitude of the electric field at a radial distance of 3.1 cm from the axis of the cylinder?

(b) What is the magnitude of the electric field at a radial distance of 5.1 cm from the axis of the cylinder?

The figure shows conducting plates with area A=1m2, and the potential on each plate. Assume you are far from the edges of the plates. •  What is the electric field between the plates in each case? •  What (and where) is the charge density on the plates in case (1)? •  What happens to an electron released midway between the plates in case (1)?

Electric potential, electric potential energy, work

In Fig. 25-39, point P is at the center of the rectangle. With V = 0 at infinity, what is the net electric potential in terms of q/d at P due to the six charged particles?

Derive an expression in terms of q2/a for the work required to set up the four-charge configuration of Fig. 25-50, assuming the charges are initially infinitely far apart.

The electric potential at points in an xy plane is given by V = (2.0 V/m2)x2 - (4.0 V/m2)y2. What are the magnitude and direction of the electric field at point (3.0 m, 3.0 m)?

Potential Energy of A System of Charges

•  4 point charges (each +Q) are connected by strings, forming a square of side L

•  If all four strings suddenly snap, what is the kinetic energy of each charge when they are very far apart?

•  Use conservation of energy: –  Final kinetic energy of all four charges

= initial potential energy stored = energy required to assemble the system of charges

+Q +Q

+Q +Q

Do this from scratch! Don’t memorize the formula in the book! We will change the numbers!!!

Potential Energy of A System of Charges: Solution

•  No energy needed to bring in first charge: U1=0

2

2 1kQU QVL

= =•  Energy needed to bring

in 2nd charge:

2 2

3 1 2( )2

kQ kQU QV Q V VL L

= = + = +

•  Energy needed to bring in 3rd charge =

•  Energy needed to bring in 4th charge =

+Q +Q

+Q +Q Total potential energy is sum of all the individual terms shown on left hand side =

2 2

4 1 2 32( )

2kQ kQU QV Q V V VL L

= = + + = +

( )242

+LkQ

So, final kinetic energy of each charge = ( )24

4

2+

LkQ

Potential V of Continuous Charge Distributions

dV =kdqr

V = dV∫Straight Line Charge: dq=λdx λ=Q/L Curved Line Charge: dq=λds λ=Q/2πR Surface Charge: dq=σdA σ=Q/πR2 dA=2πR’dR’

r = ′R 2 + z2

Potential V of Continuous Charge Distributions

Straight Line Charge: dq=λdx λ=Q/L

Curved Line Charge: dq=λds λ=Q/2πR

Potential V of Continuous Charge Distributions

Surface Charge: dq=σdA σ=Q/πR2 dA=2πR’dR‘

Straight Line Charge: dq=λdx λ=bx is given to you.

Capacitors E = σ/ε0 = q/Aε0 E = V d q = C V

Cplate = κ ε0A/d

Cplate = ε0A/d

Csphere=ε0ab/(b-a)

Connected to Battery: V=Constant Disconnected: Q=Constant

Isolated Parallel Plate Capacitor: ICPP

•  A parallel plate capacitor of capacitance C is charged using a battery.

•  Charge = Q, potential voltage difference = V. •  Battery is then disconnected. If the plate separation is INCREASED, does the

capacitance C: (a) Increase? (b) Remain the same? (c) Decrease? If the plate separation is INCREASED, does the

Voltage V: (a) Increase? (b) Remain the same? (c) Decrease?

+Q –Q

•  Q is fixed! •  d increases! •  C decreases (= ε0A/d) •  V=Q/C; V increases.

dA

VQC 0ε==

Parallel Plate Capacitor & Battery: ICPP •  A parallel plate capacitor of capacitance C is

charged using a battery.

•  Charge = Q, potential difference = V.

•  Plate separation is INCREASED while battery remains connected.

+Q –Q

•  V is fixed constant by battery! •  C decreases (=ε0A/d) •  Q=CV; Q decreases •  E = σ/ε0 = Q/ε0A decreases

Does the Electric Field Inside: (a) Increase? (b) Remain the Same? (c) Decrease?

dA

VQC 0ε==

AQE00 εε

σ ==

Battery does work on capacitor to maintain constant V!

Capacitors

Capacitors Q=CV

In series: same charge 1/Ceq= ∑1/Cj

In parallel: same voltage Ceq=∑Cj

Capacitors in Series and in Parallel

•  What’s the equivalent capacitance? •  What’s the charge in each capacitor? •  What’s the potential across each capacitor? •  What’s the charge delivered by the battery?

Capacitors: Checkpoints, Questions

Current and Resistance i = dq/dt

V = i R E = J ρ

r = ρ0(1+a(T-T0))

R = ρL/A Junction rule

Current and Resistance

Current and Resistance: Checkpoints, Questions

A cylindrical resistor of radius 5.0mm and length 2.0 cm is made of a material that has a resistivity of 3.5x10-5 Ωm. What are the (a) current density and (b) the potential difference when the energy dissipation rate in the resistor is 1.0W?