721_PartUniversity Physics Solution

721_PartUniversity Physics Solution - Capacitance and...

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Capacitance and Dielectrics 24-11 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. 24.40. IDENTIFY: Capacitance depends on geometry, and the introduction of a dielectric increases the capacitance. SET UP: For a parallel-plate capacitor, 0 /. CKA d = P EXECUTE: (a) Solving for d gives 12 2 2 3 0 9 (3.0)(8.85 10 C /N m )(0.22 m)(0.28 m) 1.64 10 m = 1.64 mm 1.0 10 F KA d C ×⋅ = × × P . Dividing this result by the thickness of a sheet of paper gives 1.64 mm 8 sheets 0.20 mm/sheet . (b) Solving for the area of the plates gives 9 2 12 2 2 0 (1.0 10 F)(0.012 m) 0.45 m (3.0)(8.85 10 C /N m ) Cd A K × = P . (c) Teflon has a smaller dielectric constant (2.1) than the posterboard, so she will need more area to achieve the same capacitance. EVALUATE: The use of dielectric makes it possible to construct reasonable-sized capacitors since the dielectric increases the capacitance by a factor of K . 24.41. IDENTIFY and SET UP: For a parallel-plate capacitor with a dielectric we can use the equation 0 d = P Minimum A means smallest possible d. d is limited by the requirement that E be less than 7 1.60 10 V/m × when V is as large as 5500 V. EXECUTE: 4 7 5500 V so 3.44 10 m V VE d d E = = × × Then 94 2 12 2 2 0 (1.25 10 F)(3.44 10 m) 0.0135 m . (3.60)(8.854 10 C / N m ) Cd A K ×× = P EVALUATE: The relation V = Ed applies with or without a dielectric present. A would have to be larger if there were no dielectric. 24.42. IDENTIFY and SET UP: Adapt the derivation of Eq.(24.1) to the situation where a dielectric is present. EXECUTE: Placing a dielectric between the plates just results in the replacement of 0 for PP in the derivation of Equation (24.20). One can follow exactly the procedure as shown for Equation (24.11). EVALUATE: The presence of the dielectric increases the energy density for a given electric field. 24.43. IDENTIFY: The permittivity P of a material is related to its dielectric constant by 0 K = . The maximum voltage is related to the maximum possible electric field before dielectric breakdown by max max d = . 0 0 E E K K P , where is the surface charge density on each plate. The induced surface charge density on the surface of the dielectric is given by i ) K . SET UP: From Table 24.2, for polystyrene 2.6 K = and the dielectric strength (maximum allowed electric field) is 7 21 0 V /m × . EXECUTE: (a) 11 2 2 (2.6) 2.3 10 C N m K = × P (b) 73 4 max max (2.0 10 V m)(2.0 10 m) 4.0 10 V d × × = × (c) 0 E K = P and 11 2 2 7 3 2 (2.3 10 C /N m )(2.0 10 V/m) 0.46 10 C/m . E == × × = × P 32 42 i 1 1 (0.46 10 C/m )(1 1/2.6) 2.8 10 C/m .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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721_PartUniversity Physics Solution - Capacitance and...

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