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Physics 2 (9 CFU) Name Surname Exercise n.1 A.A. 2011-2012 n. matricola 3.9.2012 A small sphere having mass 1g and charge Q, is hanging from a wire that forms and angle ΞΈ=15° with a vertical plate of indefinite extension bearing a surface charge density Ο=10 -8C/m2. Find the charge Q. The sphere is in equilibrium under the action of the weight force and the electrostatic force: ππ 2π0 ππ Therefore: 2π0 ππ π= β 1.7 × 10β5 πΆ π π‘π(π) = Exercise n. 2 Find an expression for the electrical resistance of a thin conducting sheet, consisting of a material of resistivity Ο, having a trapezoidal shape (see figure) with height H, thickness t, minor basis B2 and larger basis B1. Show that if B1-B2<<B2 this expression is the same as for a rectangular sheet having the same height H and basis ( π΅2+π΅1 2 ). To calculate the overall resistance we must ideally divide the sheet in a large number of rectangular resistors having height dy and width L(y), with π΅2 β π΅1 πΏ(π¦) = π¦ + π΅1 π» For the whole trapezoidal sheet the overall resistance is: π π» ππ¦ π π» ππ¦ π π» π΅2 π = β« = β« = ππ ( ) π΅2 β π΅1 π‘ 0 πΏ(π¦) π‘ 0 π΅1 π¦ + π΅1 π‘ π΅2 β π΅1 π» t is the sheet thickness. π΅2 π΅1 π΅1βπ΅2 π΅2βπ΅1 If π΅1 β π΅2 βͺ π΅2 then ππ ( ) = ππ ( β )β Therefore: π = π π» π΅1 π΅1 π΅1 π΅1 π‘ π΅1 That is, exactly the same resistance as for a rectangular sheet having the same height H and basis ( π΅2+π΅1 2 ) Exercise n. 3 A conducting square loop (side length d=10 cm) lies in the xy vertical plane, as shown in the figure. The loop has a linear mass density of 1 g cm-1. The weight force acts on the loop along the x direction. A magnetic field B is applied along the z direction. The magnetic β = πΎπ₯πΜ , K=0.1 T/cm and πΜ is the unit vector of the field varies according the expression: π΅ z axis. Calculate the current i that must flow along the loop to balance the weight force. The magnetic forces on the two sides parallel to the x axis balance out. The force on the side lying on the y axis is zero because the magnetic field is zero. The only unbalanced force acts upward on the side parallel to the y axis at a distance d, where the magnetic field is 1 T. The equilibrium conditions are: πππππ π πππ© = ππππ β π= πππ π© β ππ¨ Exercise n. 4 An electron having a kinetic energy of 10 keV moves along the x axis in a region a) of zero potential energy. At a certain point the electron hits an energy barrier of 6 keV. Calculate the wavelength associated to the electron in the region b). In the region a the wavelength is related to the kinetic energy by: β2 πΎπ = 2ππ2π In the region b we have: β2 πΎπ = πΎπ β π = 2ππ2π Therefore β ππ = β 0.16 β« β2π(πΎπ β π)