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Unformatted text preview: PHY2049 Fall 2009 Profs. A. Petkova, A. Rinzler, S. Hershfield Final Exam Solution 1. A charge +3 . 00 q lies fixed at the origin and a second charge of 2 . 00 q lies fixed in the xy plane at x = 3 . 00, y = 4 . 00. The x component of the force experienced by the 2 . 00 q charge due to the charge at the origin is? Answer: . 144 kq 2 Solution: The distance between the charges is 5 = √ 3 2 + 4 2 . The magnitude of the force between the two charges is k (2 q )(3 q ) / 5 2 . Since the 2 q charge is attracted to the +3 q charge, the xcomponent of the force on the 2 q charge is positive. To get the xcomponent we need to multiply by the cosine of the angle between the force vector and the xaxis. The cosine is the adjacent over the hypotenuse and equal to 3 / 5. Thus, the xcomponent of the force is (3 / 5) k (2 q )(3 q ) / 5 2 . 2. A circular insulating ring of radius r lies in the xy plane centered on the origin. The parts of the ring in the negative y half plane are uncharged. The parts of the ring in the positive y half plane are uniformly charged with a total charge Q . The electric field at the origin is: Answer: 2 kQ πr 2 ˆ j Solution: First, by symmetry the net electric field is in the negative ydirection if we take Q to be positive. The charge per unit length of the half ring is λ = Q/ ( πr ). Take a small element of charge, dq = λrdθ = ( Qπ ) dθ . This charge produces a field of magnitude kdq/r 2 at the origin. Letting the angle the field makes with the yaxis be θ , the ycomponent of the electric field at the origin is E y = integraldisplay π/ 2 − π/ 2 k r 2 Q π cos( θ ) dθ = kQ πr 2 (sin( π/ 2) sin( π/ 2)) . If Q is negative, then this formula still works, and the evaluated E y is in positive ydirection because then Q > 0. 3. In the figure the capacitances of the series capacitors are equal. The voltage drop across C 4 is 12 V and the charge on it is 6.0 nC. The charge on the entire network in 9.0 nC. The capacitance of C 2 is? Answer: 0.75 nF Solution: The voltage across C 1 , C 2 , and C 3 in series is 12 V. Because these capacitors are in series and equal, their effective capacitance is C 2 / 3. Since the net charge on the network is 9 nC and the charge on C 4 is 6 nC, the charge on the series resistors is 3 nC. Consequently, C 2 / 3 = 3 nC/ 12 V , which implies that C 2 = 9 / 12 nF. 4. The resistance measured between the ends of a metal wire having a circular crosssection is R. The wire diameter is halved by going through a series of rollers that preserves the volume of the wire (i.e. the wire gets correspondingly longer as its diameter shrinks with no loss of material). The resistance between the ends of the wire is now: Answer: 16R Solution: The volume of the wire is πr 2 L , where r is the radius of the wire and L is the length of the wire. If the radius is decreased by a factor of 2, then the length must be increased by a factor of 4 in order to keep the volume constant.is decreased by a factor of 2, then the length must be increased by a factor of 4 in order to keep the volume constant....
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This note was uploaded on 04/23/2010 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Charge, Force

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