PH Physics 212 Exam-2 Solution NAME: Write down your name also on the back of the package of sheets you turn in.

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1 PH Physics 212 Exam-2 Solution NAME: Write down your name also on the back of the package of sheets you turn in. SIGNATURE and ID: Return this hard copy exam together with your other answer sheets. Show all your procedures when giving the answers to the first 3 questions. For question 4 just provide the answer. 1. A solid, insulating sphere of radius a = 5 cm has a uniform charge density and a total charge Q. Concentric with this sphere there is a conducting spherical shell whose inner and outer radii are b = 2 cm and c =25 cm respectively. The electric field at a point 1 cm from the center is 3.6 x 1 3 N/C pointing radially inward, while the electric field at a point 5 cm from the center is 2 x 1 2 N/C pointing radially outward. Q 1 = - 3 C c b a P Insulator Conductor 1A (1 points) Calculate the charge Q in the insulating sphere. 1B (1 points) Calculate the charge q in on the inner surface of the shell. Calculate the charge q out on the outer surface of the shell. 1C (1 points) Calculate the electric potential difference V b - V a.

2 2. A parallel plate capacitor with plate area A = 2.5 m 2 and plate separation d = 3 mm is connected to a 45 V battery. 2A Calculate the magnitude of the charge on each plate of the capacitor. Calculate + the electric field between the plates. Calculate the energy stored in the capacitor. 45 V A = 2.5 m 2 d = 3 mm With the capacitor still connected to the battery, a slab of plastic of dielectric K = 3.6 is placed between the plates of the capacitor. 2B Calculate the magnitude of the charge on each plate of the capacitor. Calculate the electric field between the plates. Calculate the energy stored in the capacitor. 45 V K= 3.6 A = 2.5 m 2 d = 3 mm

3. The figure shows the essentials of a mass spectrometer, which can be used to measure the mass of an ion. An ion of mass m (to be measured) and charge q=1.

3 3. The figure shows the essentials of a mass spectrometer, which can be used to measure the mass of an ion. An ion of mass m (to be measured) and charge q= Coulomb is produced in the source S. The initially stationary ion is accelerated by the electric field due to a potential difference (V=1 Volts). The ion leaves S, passes through an aperture A, enters a chamber in which there exists a uniform magnetic field (B= 8 mt ) as indicated in the figure, and strikes the detector at a point that lies at x = m. 3A. (7.5 points) Calculate the electric potential difference between the points A and S (i.e. calculate V A V S ). (V A V S ) = - 1 Volts (7.5 points Calculate the electric potential energy difference of the ion between A and S (i.e. calculate U A U S ). U = q ( V) = C (- 1 Volts) = Joules 3B. (15 points) Calculate the mass m of the individual ions; give your answer in atomic mass units ( 1 u = kg). Q 1 = - 3 C A 1 Volts S

4 4. The figure below shows a parallel-plate capacitor. In what follows, assume for simplicity that the electric field produced by the plates is equivalent to the electric field produced by infinite plates charged uniformly all over their surface. In the figures below, Q stands for the initial charge accumulated in the top plate, E is the electric field in between the plates, and d is the distance between the plates. 4A (5 points) From left to right: While keeping the battery connected to the plates, an external agent (not shown in the figure) increases the distance between the plates. + E 2V - i Q i d i + 2V - E f Q f d f At t= At t > Which expression describes correctly what happens as the distance between the plates is increased? a) The electric field increases. b) The energy stored in the capacitor decreases. c) The magnitude of the charge in each plate increases. d) The electric field remains constant. e) All the expressions above are incorrect. 4B (5 points) In this case the capacitor is initially charged and then the battery is disconnected. From left to right: an external agent (not shown in the figure) increases the distance between the plates. E i Q i d i E f Q f d f At t= At t > Which expression describes correctly what happens as the distance between the plates is increased? a) The electric potential between the plates decreases.. b) The electric potential energy stored in the capacitor decreases. c) The magnitude of the charge in each plate decreases. d) The electric field remains constant. e) All the expressions above are incorrect.

9 3A. (7.5 points) Calculate the electric potential difference between the points A and S (i.e. calculate V A V S ). (V A V S ) = 1 Volts (7.5 points Calculate the electric potential energy difference of the ion between A and S (i.e. calculate U A U S ). U = q ( V) = C ( 1 Volts) = Joules 3B. (15 points) Calculate the mass m of the individual ions; give your answer in atomic mass units ( 1 u = kg).

11 Some helpful formulas: nano mm = 1 3 m Centripetal force = m v 2 / r mass of electron: 9.19 x 1 31 kg 1 ev = 1.62 x 1 19 J 1 q2 q1 1 9 Nm Coulomb's Law: F u, where r 4 C 2 Electric field, along the z axis, due to a charge Q distributed uniformly along a thin ring of radius R.: 1 Q z E kˆ 3 / z R Electric field established by an infinite and uniformly charged sheet: E = /2 o Gauss' Law = E ds = q / o, S where q is the net charge inside the Gaussian surface. Potential energy difference U B - U A W ext (A F B) ext Work done by the external force to take the particle from A to B at constant speed Definition of the Electric Potential V ( r) W ext ( q q r) Electric potential difference V B V A W ext ( A q q B) U B U A q Potential difference V V E d s B A B Electric potential due to a point charge q: A V 1 q 4 r

12 Relationship between E and V: E x = - dv/dx About capacitance Q = C V For a parallel plate capacitor C = A o /d U = CV 2 / 2 = Q 2 / 2 C Capacitors connected in parallel C equiv = C 1 + C 2 + C 3 Capacitors connected in series 1/C equiv = (1/C 1 ) + (1/C 2 ) +(1/C 3 ) sin d cos Cos d sin MAGNETISM F q v B F = force, q = charge, v = velocity, B = magnetic field 1 Tesla = 1 4 gauss q B 3 4 r v x r Magnetic field produced by a charge q that moves with velocity v = L i = Magnetic flux, L = inductance, i = current Inductive reactance X L = L B I 4 1 R B I 4 R Magnetic field at the center of a semi circle of radius "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"

13 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

HW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case.

HW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. Tipler 22.P.053 The figure below shows a portion of an infinitely