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Prepare for Exam 2 with a review of Gauss’s Law, electric potential, conductors, capacitors, and more. Practice problems included. Exam stress management tips also provided.
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Physics2113 Jonathan Dowling Lecture 20R: MON 13 OCT Exam 2: Review Session CH 23.1–9, 24.1–12, 25.1–6,8 26.1–9 Some links on exam stress: http://appl003.lsu.edu/slas/cas.nsf/$Content/Stress+Management+Tip+1 http://wso.williams.edu/orgs/peerh/stress/exams.html http://www.thecalmzone.net/Home/ExamStress.php http://www.staithes.demon.co.uk/exams.html
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.
Gauss’ law 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 = -38k N/C, and on the bottom face of the cube E = +11k N/C. Determine the net charge contained within the cube.[-2.29e-09] C
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. • What is the magnitude of the electric field at a radial distance of 3.1 cm from the axis of the cylinder? • 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)?
+Q +Q +Q +Q 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 Do this from scratch! Don’t memorize the formula in the book! We will change the numbers!!!
+Q +Q +Q +Q Potential Energy of A System of Charges: Solution • No energy needed to bring in first charge: U1=0 • Energy needed to bring in 2nd charge: • Energy needed to bring in 3rd charge = Total potential energy is sum of all the individual terms shown on left hand side = • Energy needed to bring in 4th charge = So, final kinetic energy of each charge =
Potential V of Continuous Charge Distributions 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’
Potential V of Continuous Charge Distributions Curved Line Charge: dq=λds λ=Q/2πR Straight Line Charge: dq=λdx λ=Q/L
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.
Cplate = κε0A/d Capacitors E = σ/ε0 = q/Aε0 E = V d q = C V Cplate = ε0A/d Connected to Battery: V=Constant Disconnected: Q=Constant Csphere=ε0ab/(b-a)
–Q +Q 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 is fixed! • d increases! • C decreases (= ε0A/d) • V=Q/C; V increases.
–Q +Q 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. • 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? 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?
V = i R E = Jρ Current and Resistance i = dq/dt R = ρL/A Junction rule r = ρ0(1+a(T-T0))
A cylindrical resistor of radius 5.0mm and length 2.0 cm is made of a material that has a resistivity of 3.5x10-5Ωm. What are the (a) current density and (b) the potential difference when the energy dissipation rate in the resistor is 1.0W?