oldhw1solution_pdf

# oldhw1solution_pdf - klarin(sjk772 – oldhomework 01 –...

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Unformatted text preview: klarin (sjk772) – oldhomework 01 – Turner – (56705) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Suppose that 1 . 5 g of hydrogen (H 2 ) is sep- arated into electrons and protons, and that the protons are placed at the Earth’s North Pole and the electrons are placed at the South Pole. What is the resulting compressional force on the Earth? The radius of the Earth is 6 . 37 × 10 6 m. Avogadro’s number is 6 . 02214 × 10 23 and the molar mass of the H-atom is 1 . 00782 g / mole. Correct answer: 1 . 14192 × 10 6 N. Explanation: Let : m = 1 . 5 g = 0 . 0015 kg , R E = 6 . 37 × 10 6 m , N A = 6 . 02214 × 10 23 mol − 1 , q e = 1 . 60218 × 10 − 19 C , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . The mass m is proportional to n , the num- ber of hydrogen atoms. The number of moles of hydrogen can be expressed two ways: m M H = n N A n = N A M H m = parenleftbigg 6 . 02214 × 10 23 mol − 1 1 . 00782 g / mol parenrightbigg (1 . 5 g) = 8 . 96307 × 10 23 atoms . Since each of these atoms is split into a proton and an electron, there will be n protons at the North Pole and n electrons at the South Pole, and the force is F = k e parenleftbigg nq e 2 R E parenrightbigg 2 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × bracketleftbigg 8 . 96307 × 10 23 2 (6 . 37 × 10 6 m) bracketrightbigg 2 × ( 1 . 60218 × 10 − 19 C ) 2 = 1 . 14192 × 10 6 N . 002 10.0 points The diagram shows an isolated, positive charge Q , where point B is twice as far away from Q as point A . + Q A B 10 cm 20 cm The ratio of the electric field strength at point A to the electric field strength at point B is 1. E A E B = 1 1 . 2. E A E B = 1 2 . 3. E A E B = 4 1 . correct 4. E A E B = 2 1 . 5. E A E B = 8 1 . Explanation: Let : r B = 2 r A . The electric field strength E ∝ 1 r 2 , so E A E B = 1 r 2 A 1 r 2 B = r 2 B r 2 A = (2 r ) 2 r 2 = 4 . 003 10.0 points Three point charges are located at the vertices of an equilateral triangle. The charge at the top vertex of the triangle is given in the figure. The two charges q at the bottom vertices of the triangle are equal. A fourth charge 3 μ C klarin (sjk772) – oldhomework 01 – Turner – (56705) 2 is placed below the triangle on its symmetry- axis, and experiences a zero net force from the other three charges, as shown in the figure below. 3 . 6 m 1 . 2m − 6 . 2 μ C q q 3 μ C Find q . Correct answer: 1 . 40296 μ C. Explanation: Let : Q 1 = − 6 . 2 μ C , Q 2 = q , Q 3 = q , Q 4 = 3 μ C , a = 3 . 6 m , and d = − 1 . 2 m . a d Q 1 Q 2 Q 3 3 μ C θ The force on Q 4 is due to the Coulomb forces from Q 1 , Q 2 , and Q 3 . Because Q 2 and Q 3 have equal charge, the x-components of their forces cancel out (by symmetry). Thus we only need to consider the y-components of the forces....
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## This note was uploaded on 10/10/2010 for the course PHY 56705 taught by Professor Turner during the Fall '10 term at University of Texas.

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oldhw1solution_pdf - klarin(sjk772 – oldhomework 01 –...

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