oldhw1solution_pdf - klarin (sjk772) oldhomework 01 Turner...

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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 Earths 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. Avogadros 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 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 C q q 3 C Find q . Correct answer: 1 . 40296 C. Explanation: Let : Q 1 = 6 . 2 C , Q 2 = q , Q 3 = q , Q 4 = 3 C , a = 3 . 6 m , and d = 1 . 2 m . a d Q 1 Q 2 Q 3 3 C 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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oldhw1solution_pdf - klarin (sjk772) oldhomework 01 Turner...

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