Final_solution

Final_solution - Physics 121 Honors Final Exam Spring 2003...

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Unformatted text preview: Physics 121 Honors Final Exam Spring 2003 Name: Section: Closed book exam. Only one 8 . 5 00 × 11 00 formula sheet (front and back side) can be used. Calculators are allowed. Use the scantron forms (pencil only!) for the multiple choice problems. Circle the answers on the examination sheet as well, and return it together with the scantron form. Use the back of these pages, or attach your own pages with solutions for problems that require calculations. The 14 multiple-choice problems are 3 . 6 points each. Partial credit (up to 2 points) will be given for multiple-choice problems, if you provide a detailed solution on the examination sheets. The six work-out problems are 10 points each. The nominal number of points is 100. Clearly print your first and last name and indicate your section number on both the scantron form and the examination sheet. Good luck! Multiple-Choice Problems Problem 1: The electrons in a particle beam each have a kinetic energy of 1.25 keV. What is the magnitude of the electric field that stops these electrons in a distance of 2.8 cm? A) 1 . 75 × 10 1 V/m B) 2 . 11 × 10 2 V/m C) 1 . 26 × 10 3 V/m D) 4 . 46 × 10 4 V/m E) 8 . 93 × 10 4 V/m K = 1 2 mv 2 ⇒ v 2 = 2 K m v 2 = v 2 + 2 a ( x- x ) and v = ⇒ a = v 2 2 d = K md F net = qE = ma ⇒ E = ma e = K ed = 4 . 46 × 10 4 V / m Problem 2: A thin glass rod is bend into a semicircle of radius r = 14 . 0 cm. A charge of + 7 . 50 μ C is uniformly distributed along the upper half, and a charge of- 7 . 50 μ C is uniformly distributed along the lower half. Find the magnitude and direction of the electric field ~ E at point P , the center of the semicircle! A) 8 . 67 × 10 5 V/m (down) B) 4 . 38 × 10 6 V/m (down) C) 6 . 20 × 10 6 V/m (down) D) 1 . 24 × 10 7 V/m (left) E) 1 . 95 × 10 7 V/m (up) dq = λ ds ⇒ dE r = dE cos θ = 1 4 πε λ r 2 cos θ ds ds = rd θ ⇒ E r = Z + 45 ◦- 45 ◦ 1 4 πε λ r 2 cos θ rd θ = λ 4 πε r h sin θ i + 45 ◦- 45 ◦ = √ 2 ( 4 q / 2 π r ) 4 πε r = q √ 2 π 2 ε r 2 ~ E = ~ E + + ~ E- =- √ 2 E r ˆ j =- ( 4 . 38 × 10 6 V / m ) ˆ j Problem 3: A point charge q p = + q is placed inside a spherical conducting shell with inner radius a and outer radius b . The total charge of the conductor is Q =- 4 q . What are the surface charges on the inner surface q a and outer surface q b ? A) q a =- q and q b =- 3 q B) q a =- 3 q and q b =- 1 q C) q a = 0 and q b =- 4 q D) q a = + q and q b =- 4 q E) q a =- q and q b =- 4 q q a =- q p =- q q b = Q- q a =- 4 q- (- q ) =- 3 q Problem 4: Two protons are fixed 28.0 mm apart ( m p = 1 . 67 × 10- 27 kg). Another proton is shot from infinity ( V ∞ = 0!) and stops midway between the two protons. What is the protons initial speed?...
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This note was uploaded on 04/18/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.

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Final_solution - Physics 121 Honors Final Exam Spring 2003...

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