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 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 x-axis 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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extra_credit 01-solutions - Reilly(mar3978 – extra credit...

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