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Unformatted text preview: Reilly (mar3978) – extra credit 01 – turner – (56725) 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points How much positive charge is in 1 . 5 kg of hy drogen? The charge on a proton is 1 . 60218 × 10 − 19 C and Avogadro’s number is 6 . 023 × 10 23 . The atomic weight (1 . 00794 g) of hydrogen contains Avogadro’s number of atoms, with each atom having 1 protons and 1 electrons. Correct answer: 1 . 43608 × 10 8 C. Explanation: Let : e = 1 . 60218 × 10 − 19 C , m = 1 . 5 kg = 1500 g , N A = 6 . 023 × 10 23 , AW C = 1 . 00794 g , and n pr = 1 protons / atom . N pr = m m A n pr = mn pr N A AW = (1500 g) (6 . 023 × 10 23 atoms) 1 . 00794 g × (1 protons / atom) = 8 . 96333 × 10 26 protons and Q = eN A = (1 . 60218 × 10 − 19 C) (8 . 96333 × 10 26 ) = 1 . 43608 × 10 8 C . 002 10.0 points A line of charge starts at x = x , where x is positive, and extends along the xaxis to positive infinity. If the linear charge den sity is given by λ = λ x /x , where λ is a constant, determine the electric field at the origin. (Here ˆ ı denotes the unit vector in the positive x direction.) 1. k λ 2 x (ˆ ı ) 2. k λ x (ˆ ı ) 3. k λ 2 2 x (ˆ ı ) 4. k λ 2 x 2 (ˆ ı ) 5. k λ x ( ˆ ı ) 6. k λ 2 x ( ˆ ı ) correct Explanation: First we realize that we are dealing with a continuous distribution of charge (as opposed to point charges). We must divide the dis tribution into small elements and integrate. Using Coulomb’s law, the electric field cre ated by each small element with charge dq is dE = k dq x 2 where dq = λdx = λ x x dx. Now we integrate over the entire distribution (i.e. from x = x to x = + ∞ ) and insert our dq :  vector E  = integraldisplay k dq x 2 = integraldisplay ∞ x k λ x dx x 3 = k λ x 2 1 x 2 vextendsingle vextendsingle vextendsingle ∞ x = k λ 2 x . Since the distribution is to the right of the point of interest, the electric field is directed along the x axis if λ is positive. That is, a positive charge at the origin would experience a force in the direction of ˆ ı from this charge distribution. In fact, the direction of an elec tric field at a point P in space is defined as the direction in which the electric force acting on a positive particle at that point P would point. So vector E = k λ 2 x ( ˆ ı ) . Reilly (mar3978) – extra credit 01 – turner – (56725) 2 003 10.0 points Two metal spheres that are initially un charged are mounted on insulating stands, as shown....
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 Spring '10
 TURNER,J
 Charge, Electric charge, Reilly

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