Old Exam 2 Sol

# Old Exam 2 Sol - Platt David – Oldquiz 2 – Due 4:00 am...

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Unformatted text preview: Platt, David – Oldquiz 2 – Due: Oct 16 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points Consider a long coaxial arrangement with a cylindrical wire of radius a along the axis of a thin cylindrical shell of radius b . There is a charge of- Q on the inner wire and charge of + Q on the outer shell. The figure below shows a short segment (length ‘ ) of the coaxial cable. Assume the length is much greater than the radii of the cylinders ( ‘ À b ). ‘ + Q , b P , r- Q , a The magnitude of the electric field at P where P is at a radius r , between the wire and the shell, is given by 1. k ~ E k = Q 2 π ² r 2 . 2. k ~ E k = Q 2 4 π ² r 2 . 3. k ~ E k = Q 2 π r‘ . 4. k ~ E k = Q 2 π ² r ‘ . correct 5. None of these 6. k ~ E k = Q 4 π r 2 . 7. k ~ E k = Q 2 2 π ² r ‘ . 8. k ~ E k = Q 2 2 π r‘ . 9. k ~ E k = Q 2 4 π r 2 . Explanation: Construct a Gaussian cylinder at radius r (for a < r < b ). The charge enclosed is the charge on the wire, + Q . Thus, Φ = I ~ E · d ~ A = Q ² E 2 π r ‘ = Q ² ˆ r k ~ E k = Q 2 π ² r ‘ . In the second step the following reasoning was used. The total flux coming out from the cylindrical Gaussian surface is the sum of φ ends , the flux coming out from the ends of the cylinder and φ side the flux coming out from the side of the cylinder. For ‘ large compared with a and b we can neglect φ ends and take the electric field to be radial, so φ = φ side = E 2 π r ‘ . Since ~ E points radially, ~ E = Q 2 π ² r ‘ ˆ r . 002 (part 2 of 3) 10 points The magnitude of the potential difference be- tween the two conductors is given by 1. | Δ V | = Q 4 π ² µ 1 a- 1 b ¶ . 2. | Δ V | = Q 2 2 π ² ‘ ln µ b a ¶ . 3. | Δ V | = Q 2 2 π ‘ ln µ b a ¶ . 4. | Δ V | = Q 4 π µ 1 a- 1 b ¶ . Platt, David – Oldquiz 2 – Due: Oct 16 2005, 4:00 am – Inst: Ken Shih 2 5. | Δ V | = Q 2 π ² ‘ ln µ b a ¶ . correct 6. None of these 7. | Δ V | = Q 2 π ‘ ln µ b a ¶ . 8. | Δ V | = Q 2 4 π µ 1 a- 1 b ¶ . 9. | Δ V | = Q 2 4 π ² µ 1 a- 1 b ¶ . Explanation: The absolute value of the potential differ- ence is given by the expression Δ V =- Z ~ E · d ~ ‘. This is a line integral along a radius vector from r = a to r = b . Thus, d ~ ‘ = dr ˆ r. | Δ V | = Z b a Q 2 π ² r ‘ dr ˆ r · ˆ r . Since ˆ r is a unit vector, ˆ r · ˆ r = 1 . Thus, | Δ V | = Q 2 π ² ‘ Z b a dr r = Q 2 π ² ‘ ln µ b a ¶ . 003 (part 3 of 3) 10 points Given: The capacitance C , the length ‘ , and radius of the cylindrical shell b as shown in the figure....
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Old Exam 2 Sol - Platt David – Oldquiz 2 – Due 4:00 am...

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