Physics9C_A_HW5_Soln

# Physics9C_A_HW5_Soln - CAPACITANCE AND DIELECTRICS 24.18....

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24-1 C APACITANCE AND D IELECTRICS 24.18. IDENTIFY: For capacitors in parallel the voltages are the same and the charges add. For capacitors in series, the charges are the same and the voltages add. / CQV = . SET UP: 1 C and 2 C are in parallel and 3 C is in series with the parallel combination of 1 C and 2 C . EXECUTE: (a) 12 and CC are in parallel and so have the same potential across them: 6 2 6 2 40.0 10 C 13.33 V 3.00 10 F Q VV C × == = = × . Therefore, 66 11 1 (13.33 V)(3.00 10 F) 80.0 10 C QV C == × . Since 3 C is in series with the parallel combination of and , its charge must be equal to their combined charge: 6 3 40.0 10 C 80.0 10 C 120.0 10 C C −− = × . (b) The total capacitance is found from tot 12 3 1 1 1 9.00 10 F 5.00 10 F CCC =+ = + ×× and tot 3.21 F C µ = . 6 tot 6 tot 120.0 10 C 37.4 V 3.21 10 F ab Q V C × = × . EVALUATE: 6 3 3 6 3 120.0 10 C 24.0 V Q V C × = × . 13 ab VVV = + . 24.28. IDENTIFY: After the two capacitors are connected they must have equal potential difference, and their combined charge must add up to the original charge. SET UP: / = . The stored energy is 2 2 1 22 Q UC V C EXECUTE: (a) 0 . QC V = (b) QQ V and also 0 QQ QC V +== . 1 = and 2 2 C C = so (2 ) = and 1 2 2 Q Q = . 1 3 2 = . 1 2 3 = and 1 0 33 . (c) 2 21 2 1233 0 () 2 () 1 1 3 3 Q V C C C   = + = =     (d) The original U was 2 1 0 2 V = , so 2 0 1 6 V ∆= . (e) Thermal energy of capacitor, wires, etc., and electromagnetic radiation. EVALUATE: The original charge of the charged capacitor must distribute between the two capacitors to make the potential the same across each capacitor. The voltage V for each after they are connected is less than the original voltage 0 V of the charged capacitor. 24.37. IDENTIFY: Use the rules for series and for parallel capacitors to express the voltage for each capacitor in terms of the applied voltage. Express U, Q, and E in terms of the capacitor voltage. SET UP: Le the applied voltage be V. Let each capacitor have capacitance C. 2 1 2 V = for a single capacitor with voltage V. EXECUTE: (a) series Voltage across each capacitor is /2. V The total energy stored is ( ) s 24 2[ / 2 ] V C V parallel Voltage across each capacitor is V . The total energy stored is ( ) 1 p 2 2 V C V p s 4 UU = 24

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24-2 Chapter 24 (b) QC V = for a single capacitor with voltage V . ( ) sp p s 2[ / 2 ] ; 2 ( )2 ; Q 2 V C V V C V Q == = = = (c) / E Vd = for a capacitor with voltage V. p s /2 ; / ; 2 E VdE V dE E = EVALUATE: The parallel combination stores more energy and more charge since the voltage for each capacitor is larger for parallel. More energy stored and larger voltage for parallel means larger electric field in the parallel case. 24.39. IDENTIFY and SET UP: Q is constant so we can apply Eq.(24.14). The charge density on each surface of the dielectric is given by Eq.(24.16). EXECUTE: 5 00 5 3.20 10 V/m so 1.28 2.50 10 V/m EE EK KE × = = × (a) (1 1/ ) i K σσ =− 12 2 2 5 6 2 (8.854 10 C /N m )(3.20 10 N/C) 2.833 10 C/m E σ −− × × = × P 62 72 (2.833 10 C/m )(1 1/1.28) 6.20 10 C/m i = × (b) As calculated above, 1.28. K = EVALUATE: The surface charges on the dielectric produce an electric field that partially cancels the electric field produced by the charges on the capacitor plates.
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## Physics9C_A_HW5_Soln - CAPACITANCE AND DIELECTRICS 24.18....

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