extra_credit 01-solutions

extra_credit 01-solutions - Reilly (mar3978) extra credit...

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Unformatted text preview: Reilly (mar3978) extra credit 01 turner (56725) 1 This print-out should have 9 questions. Multiple-choice 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 Avogadros number is 6 . 023 10 23 . The atomic weight (1 . 00794 g) of hydrogen contains Avogadros 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 x-axis to positive infinity. If the linear charge den- sity is given by = x /x , where is a constant, determine the electric field at the origin. (Here denotes the unit vector in the positive x direction.) 1. k 2 x ( ) 2. k x ( ) 3. k 2 2 x ( ) 4. k 2 x 2 ( ) 5. k x (- ) 6. k 2 x (- ) 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 Coulombs law, the electric field cre- ated by each small element with charge dq is dE = k dq x 2 where dq = dx = 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 x dx x 3 =- k x 2 1 x 2 vextendsingle vextendsingle vextendsingle x = k 2 x . Since the distribution is to the right of the point of interest, the electric field is directed along the- x axis if is positive. That is, a positive charge at the origin would experience a force in the direction of- 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 2 x (- ) . 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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extra_credit 01-solutions - Reilly (mar3978) extra credit...

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