PH Physics 211 Exam-2
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1 PH Physics 211 Exam2 NAME: Write down your name also on the back of your exam. SIGNATURE and ID: Write down your name on the back of the package of sheets you turn in. Notice: The exam is worth 11 points: Q1= 3 points, Q2= 4 points, and Q3= 4 points. (i.e. you have an opportunity to earn extra 1 point in this exam). 1. 1A The figure at the right shows portions of two infinitely large, parallel, nonconducting sheets, each with a fixed uniform charge. The surface charge densities are 1 = 6 C/m 2 and 2 = 2 C/m 2. P S U (7.5 points) Three points P, S, and U are located at the vertices of a equilateral triangle of side 1 mm. Rank the points according to the electrical potential at those positions respectively, greatest first. (Circle your answer). a) U, P, S b) P, U, S c) P, S, U d) U, S, P e) NA (7.5 points) Three electrons are located at the vertices of a equilateral triangle PSU of side 1 mm. Rank the electrons according to their electrical potential energy, greatest first. (Circle your answer). a) U, P, S b) P, U, S c) P, S, U d) U, S, P e) NA
2 1B (7.5 points) The figure shows three situations in which a positively charged particle moves at velocity v through a uniform magnetic field B and experiences a magnetic force. In each situation, determine whether the orientation of the vectors are physically reasonable. (Circle your answer). x B v. B v F B. v F B F B B I II III a) F, F, T b) T, F, T c) T, T, F d) T, F, F e) NA 1C (7.5 points) The figure shows four directions for the velocity vector v of a negatively charged particle moving through a uniform magnetic field B pointing to the left. (Of course, the magnetic fields exist all over the space). Rank the directions according to the magnitude of the net force on the particle, greatest first. (Circle your answer). B a) 1,2,3,4 b) 1 tie with 4, 2, 3 d) 2, 3 tie with 4, 1 c) 3, 2, 1 tie with 4 e) NA
3 2. A total charge of Q = 4 x 1 6 Coulomb is uniformly distributed all over the volume of a nonconducting sphere of radius R= 3 cm. The electric field is radially directed and has a magnitude equal to: 1 Q E( r) r for r < R 3 4 R E ( r ) 1 Q 2 4 r for r > R The potential is chosen to be zero at r=r, i.e. V(R) = Volts: 2A (12 points) Find general expression for the potential V(r), for the range r <R. (i.e. give an expression for the potential V as a function of r that satisfies the condition V(R) = Volts ). Note: Provide all your detailed procedure leading to your answer. Writing directly just the answer will receive zero points in the grading). 2B (12 points) Find general expression for the potential V(r), for the range r > R. (i.e. give an expression for the potential V as a function of r that satisfies the condition V(R) = Volts). Note: Provide all your detailed procedure leading to your answer. Writing directly just the answer will receive zero points in the grading). 2C (12 points) Calculate the potential at r =; i.e. calculate V(). Calculate the potential at infinity, i.e. calculate V( ). 2D (4 points) Sketch the potential V as a function of r, from r= to r =. f Hint: Use the expression for the potentialdifference E r V V d, and capitalize on the fact that you can choose arbitrarily what point constitutes the initial point and what constitutes the final point. I would suggest to choose the initial and final points in such a way that the electric field E and the displacement d r point in the same direction. To double check that the signs ( or ) in the calculations above are correct, you may want to verify them by using the alternative definition of potentialdifference based on the work done by an external force: Wext ( A q B) V ( B) V ( A) (where A and B can be chosen arbitrarily as q the initial or final positions, respectively) f i i
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7 3. The figure shows a parallel plate capacitor with a plate area A = 24 (mm) 2 and a separation distance d = 1 mm. The left half of the gap is filled with material of dielectric constant K 1 =3.14 ; the right half is filled with material of dielectric constant K 2 = The capacitor is connected to a 8 V battery. 8 Volts K K d= 1 mm Side view 3A (12 points) Calculate the electric field on the lefthalf of the capacitor. Calculate the electric field on the righthalf of the capacitor 3B (4 points) Sketch a possible charge distribution on each plate. 3B (12 points) Calculate the surface charge density on the lefthalf of the plates. Calculate the surface charge density on the righthalf of the plates. 3C (12 points) Calculate the capacitance of the parallelplate capacitor.
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11 Some formulas: nano ln( ab ) = ln( a ) ln ( b ) ln( a / b ) = ln( a ) ln ( b ) Electron mass: 9.1 x 1 31 Kg Proton mass = 1.67 x 1 27 Kg Electron charge 1.6 x 1 19 C 1 Gauss = 1 4 Tesla 2 v Centripetal acceleration: a c R 1eV = 1.6 x 1 19 J o = 8.85 x 1 12 C 2 /N.m 2, o = 9 x 1 9 N.m 2 /C 2, o/= 1 7 T m/a 1 2 sin d cos, Cos d sin, sin cos d sin 2 ELECTRICITY 1 q2 q1 Coulomb's Law: F uˆ 2 r 4 r Electric filed for an infinite and uniformlycharged sheet: E = /2 o is the surface charge density. For a uniformly charged ring with radius R and total charge q: 1 q z E( z) (Field along the axis passing through the center of the ring) 2 2 3/ 2 4 ( z R ) Gauss' Law = E ds = q / o, where q is the net charge inside the S Gaussian surface S Definition of Electric Potential V ( r) W ext ( q q r)
12 Assuming the potential at infinity is zero Definition of Electric Potential difference Electric potential due to a point charge q: (Assuming the potential at infinite is zero) Wext ( B q V ( A) V ( B) q 1 q V ( r) 4 r A) V Potential difference : V f i i f E dr Relationship between E and V: E x = dv/dx About capacitance Q = C V U = CV 2 / 2 = Q 2 / 2 C For a capacitor of parallel plates: C = A o /d RC circuit: Time constant = RC MAGNETISM F = q v x B F = force, q= charge, v = velocity, B = magnetic field Magnetic field produced by a charge q that moves with velocity v q B 3 4 r v x r = L i = Magnetic flux, L = inductance, i = current Hall effect BI = nqtv Hall Inductive reactance X L = L B I 4 1 R Magnetic field at the center of a semicircle of radius "R"
13 B I 4 R Magnetic field at the center of an arc of angle f (in radians) and radius "R". B I 2 r Magnetic field produced by a infinitely long wire at a distance "r" from it. F L I a I b 2 d Force per unit length between two parallel long wires, carrying currents Ia and Ib respectively, separated by a distance "d" Faraday's Law, t where = Magnetic flux and = electromotive force Definition of the magnetic dipole moment of a loop of area A, carrying a current I: = I A n where A = area, I current, n = unit vector perpendicular to the loop
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