303L-HW2 - fierro (jmf2547) – HW 02 – berk – (60290)...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: fierro (jmf2547) – HW 02 – berk – (60290) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points You have 2 . 6 kg of water. One mole of water has a mass of 18 . 1 g / mol and each molecule of water contains 10 electrons since water is H 2 O. What is the total electron charge contained in this volume of water? Correct answer: − 1 . 38409 × 10 8 C. Explanation: Let : N A = 6 . 02214 × 10 23 molec / mol , q e = − 1 . 6 × 10- 19 C / electron , m = 2 . 6 kg , M = 18 . 1 g / mol = 0 . 0181 kg / mol , and Z = 10 electrons / molec . The mass is proportional to the number of molecules, so for m grams in n molecules and M grams in N A molecules, m M = n N A n = m M N A Since 10 electrons are in each molecule of water, then the total number of electrons n e in the coin is n e = Z n = Z m M N A and the total charge q for the n e electrons is q = n e q e = Z m N A q e M = (10 electrons / molec) 2 . 6 kg . 0181 kg / mol × ( 6 . 02214 × 10 23 molec / mol ) × ( − 1 . 6 × 10- 19 C / electron ) = − 1 . 38409 × 10 8 C . 002 10.0 points A charge Q is spread uniformly along the circumference of a circle of radius R . A point charge q is placed at the center of this circle. What is the total force exerted on q as calculated by Coulomb’s law? 1. Use R for the distance. 2. The result of the calculation is zero. cor- rect 3. Use 2 π R for the distance. 4. Use 2 R for the distance. 5. None of these. Explanation: By symmetry and using the fact that the charge is uniformly distributed along the cir- cumference, the total force on q is zero. For example, consider a small element of charge δQ on the circle. The force on q is along the vector connecting δQ and q . But on the ex- act opposite side there is another element of charge δQ which exerts an equal but opposite force on q . This is true for every point on the circle, so the net force is zero. 003 (part 1 of 2) 10.0 points Two conducting spheres have identical radii. Initially they have charges of opposite sign and unequal magnitudes with the magnitude of the positive charge larger than the mag- nitude of the negative charge. They attract each other with a force of 0 . 123 N when sepa- rated by 0 . 6 m. + + + + − − Initial The spheres are suddenly connected by a thin conducting wire, which is then removed. + + Connected fierro (jmf2547) – HW 02 – berk – (60290) 2 Now the spheres repel each other with a force of 0 . 03 N....
View Full Document

This note was uploaded on 09/19/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

Page1 / 6

303L-HW2 - fierro (jmf2547) – HW 02 – berk – (60290)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online