HW6solutions

# HW6solutions - Physics 213 HW#6 – Solutions Spring 2008...

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Unformatted text preview: Physics 213 HW #6 – Solutions Spring 2008 24.14. IDENTIFY: The capacitors between b and c are in parallel. This combination is in series with the 15 pF capacitor. SET UP: Let C 1 = 15 pF , C 2 = 9.0 pF and C 3 = 11 pF . EXECUTE: (a) For capacitors in parallel, C eq = C 1 + C 2 + ¡ so C 23 = C 2 + C 3 = 20 pF (b) C 1 = 15 pF is in series with C 23 = 20 pF . For capacitors in series, 1 C eq = 1 C 1 + 1 C 2 + ¡ so 1 C 123 = 1 C 1 + 1 C 23 and C 123 = C 1 C 23 C 1 + C 23 = (15 pF)(20 pF) 15 pF + 20 pF = 8.6 pF . EVALUATE: For capacitors in parallel the equivalent capacitance is larger than any of the individual capacitors. For capacitors in series the equivalent capacitance is smaller than any of the individual capacitors. 24.60. IDENTIFY: Apply the rules for combining capacitors in series and in parallel. SET UP: With the switch open each pair (top & bottom) of 3.00 μ F and 6.00 μ F capacitors are in series with each other and each pair is in parallel with the other pair. When the switch is closed each pair (left and right) of 3.00 μ F and 6.00 μ F capacitors are in parallel with each other and the two pairs are in series. EXECUTE: (a) With S open, C eq = 1 3 μ F + 1 6 μ F ¡ ¢ £ ¤ ¥ ¦ § 1 + 1 3 μ F + 1 6 μ F ¡ ¢ £ ¤ ¥ ¦ § 1 = 4 μ F and Q tot = C eq V = 4 μ F ¡ 210 V = 840 μ C . By symmetry, each capacitor carries Q tot /2, or 420 μ C. The voltages are then calculated via V = Q / C . This gives V ad = Q 3 / C 3 = 140 V and V ac = Q 6 / C 6 = 70 V . V cd = V ad ¡ V ac = 70 V . (b) When S is closed, by symmetry, each capacitor has the same potential difference of 105 V. [Also note: points c and d must be at the same potential, and C eq = 1 (3 + 6) μ F + 1 (3 + 6) μ F ¡ ¢ £ ¤ ¥ ¦ § 1 = 4.5 μ F . Q tot = C eq V = 4.5 μ F ¡ 210V = 945 μ C ]. (c) V 3 = V 6 = Q 3 C 3 = Q 6 C 6 , and since C 6 =2C 3 , Q 6 =Q 3 /2. Now Q 3 +Q 6 =3Q 6 =Q tot , so Q 6 =Q tot /3=315 μ C and Q 3 =630 μ C. The only way for the sum of the + charge on one plate of C 6 and the - charge on one plate of C 3 to change is for charge to flow through the switch, with quantity equal to the change in Q 3 ¡ Q 6 . With the switch open, Q 3 = Q 6 and Q 3 ¡ Q 6 = 0....
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## This note was uploaded on 08/27/2008 for the course PHYS 2213 taught by Professor Perelstein,m during the Spring '07 term at Cornell.

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HW6solutions - Physics 213 HW#6 – Solutions Spring 2008...

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