exam 2: review session ch 23.1–9, 24.1–12, 25.1–6,8...
TRANSCRIPT
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?