HW 8 solution

# HW 8 solution - song(shs546 – Homework 8 – weathers...

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Unformatted text preview: song (shs546) – Homework 8 – weathers – (22202) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points When a certain air-filled parallel-plate ca- pacitor is connected across a battery, it ac- quires a charge (on each plate) of magnitude 163 μ C. While the battery connection is maintained, a dielectric slab is inserted into the space between the capacitor plates and completely fills this region. This results in the accumulation of an additional charge of 266 μ C on each plate. κ κ What is the dielectric constant of the slab? Correct answer: 2 . 6319. Explanation: Let : Q a = 163 μ C , Q d = 163 μ C + 266 μ C , and = 429 μ C . Since the capacitor is connected to the bat- tery the whole time, we know that the poten- tial drop across the capacitor is held constant. The charge changes only because the capaci- tance does. When the capacitor is filled with air, we have Q a = C a V . When the dielec- tric slab is inserted, the charge is given by Q d = C d V . The dielectric constant is then κ = C d C a = Q d Q a = 429 μ C 163 μ C = 2 . 6319 . 002 (part 1 of 4) 10.0 points Four capacitors are connected as shown in the figure. 1 6 . 2 μ F 5 7 . 1 μ F 8 8 . 2 μ F 98 . 2 V 46 . 8 μ F a b c d Find the capacitance between points a and b of the entire capacitor network. Correct answer: 118 . 153 μ F. Explanation: Let : C 1 = 16 . 2 μ F , C 2 = 46 . 8 μ F , C 3 = 57 . 1 μ F , C 4 = 88 . 2 μ F , and E B = 98 . 2 V . C 1 C 3 C 4 E B C 2 a b c d A good rule of thumb is to eliminate junc- tions connected by zero capacitance. C 2 C 3 C 1 C 4 a b c d The definition of capacitance is C ≡ Q V . The parallel connection of C 1 and C 2 gives the equivalent capacitance C 12 = C 1 + C 2 = 16 . 2 μ F + 46 . 8 μ F = 63 μ F . song (shs546) – Homework 8 – weathers – (22202) 2 C 12 C 3 C 4 a b The series connection of C 12 and C 3 gives the equivalent capacitance C 123 = 1 1 C 12 + 1 C 3 = C 12 C 3 C 12 + C 3 = (63 μ F) (57 . 1 μ F) 63 μ F + 57 . 1 μ F = 29 . 9525 μ F . C 123 C 4 a b The parallel connection of C 123 and C 4 gives the equivalent capacitance C ab = C 123 + C 4 = 29 . 9525 μ F + 88 . 2 μ F = 118 . 153 μ F , or combining the above steps, the equivalent capacitance is C ab = ( C 1 + C 2 ) C 3 C 1 + C 2 + C 3 + C 4 = (16 . 2 μ F + 46 . 8 μ F) (57 . 1 μ F) 16 . 2 μ F + 46 . 8 μ F + 57 . 1 μ F + 88 . 2 μ F = 118 . 153 μ F ....
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## This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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HW 8 solution - song(shs546 – Homework 8 – weathers...

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