P2213_HW05_solns_Fall2008

P2213_HW05_solns_Fall2008 - Physics 2213 HW#5 Solutions...

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- 1 - Physics 2213 HW #5 — Solutions Fall 2008 21.8 [Aluminum Spheres] We’ll use the mass of a sphere and the atomic mass of aluminum to find the number of aluminum atoms in one sphere. Each atom has 13 electrons. Then, we’ll apply Coulomb’s law and calculate the magnitude of charge on each sphere. . , where is the number of electrons removed from one sphere and added to the other. (a) The total number of electrons on each sphere equals the number of protons. . (b) For a force of N to act between the spheres, . This gives q = 4 πε 0 (1.00 × 10 4 N)(0.800 m) 2 = 8.43 × 10 4 C . The number of electrons is (c) . Summary: When ordinary objects receive a net charge the fractional change in the total number of electrons in the object is very small . 21.49 [Electric Field in 2-D] The electric field of a positive charge is directed radially outward from the charge and has magnitude E = 1 4 πε 0 q r 2 = k q r 2 . The resultant electric field is the vector sum of the fields of the individual charges. The placement of the charges is shown in the figure at right. Point a is symmetrically placed between identical charges, so symmetry tells us the electric field must be zero. Point b is to the right of both charges and both electric fields are in the + x -direction and the resultant field is in this direction. At point c both fields have a downward component and the field of has a component to the right, so the net is +x, -y. At point d both fields have an upward component but by symmetry they have equal and opposite x
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