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Unformatted text preview: Prof Liphardt Physics 7B Midterm 2 Sp 2010 Problems 0, 1, and 2 Brian Henning April 13, 2010 Problem 0  10 points Consider the diagram shown. For each option below, state whether it is true or false. There is only one correct option NOTE: I graded the problem only on true/false, not on whether you showed why each state ment was true or false. That said, I saw many times wrong reasoning lead to the correct answer of true/false, so I believe it is beneficial to carefully go through the solutions. (a) The electric field at point O is q/ 8 πǫ R 2 FALSE: By symmetry, the electric fields due to points A and B cancel and the only contribu tion to the electric field at the origin is due to point C (see diagram). vector E ( O ) = vector E c ( O ) = 1 4 πǫ q c R 2 = − 1 6 πǫ q R 2 negationslash = − 1 8 πǫ q R 2 (b) The potential energy of the system is zero FALSE: The potential energy of the configuration is given by summing over the potential energy of pairs of charges: U = summationdisplay pairs k q i q j l ij = k parenleftbigg q B q C l BC + q A q C l AC + q A q B l AB parenrightbigg 1 Figure 1: The electric field at the origin from A and B cancel by symmetry. The only component is the vector Efield due to C, which points to the left. After a little trigonometry we find l BC = √ 3 R,l AC = R,l AB = 2 R so the potential energy is: U = k q 2 9 R parenleftbigg − 2 √ 3 − 2 1 + 1 2 parenrightbigg negationslash = 0 Note that since all the charges sum to zero ( ∑ q i = 0), if the charges were equidistant ( l AC = l AB = l BC ) then the potential energy would be zero and the charges would hold themselves in place. For the given configuration, if the charges were not held in place they would collapse, and this is reflected in the nonzero potential energy of the configuration. (c) The magnitude of the force between the charges at C and B is q 2 / 54 πǫ R 2 TRUE: Use Coulomb’s law to verify: vextendsingle vextendsingle vextendsingle vector F BC vextendsingle vextendsingle vextendsingle = k q c q B l 2 BC = 1 4 πǫ 2 9 q 2 3 R 2 = 1 54 πǫ q 2 R 2 (d) The potential at point O is q/ 12 πǫ R FALSE: The potential at the origin is actually zero. The electric potential is the sum of the potential from each individual charge by superposition, V = ∑ i V i . Take the potential to be zero at infinity so that the potential of each charge is given by V i = k q i r i where r i is the distance from the charge q i to the point of interest (in this case, O). Since all charges are equidistant to the origin and there charges sum to zero, the potential at the origin is zero: V ( O ) = k summationdisplay i q i R = k R summationdisplay i q i = 0 Problem 2  15 points A ring of charge with radius R has a gap of arc length s (see Fig ). What is the electric field in the center (magnitude and direction) if the ring caries a uniformly distributed charge of...
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This note was uploaded on 06/30/2010 for the course PHYSICS ?? taught by Professor Zettle during the Spring '08 term at Berkeley.
 Spring '08
 ZETTLE

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