Fisica 2 - Cap 26 7 Edição

44. (a) The electric field E3 in the free space between the two plates is E3 = q/e0d while that inside the slab is E2 = E3/K = q/Ke0d. Thus, V3 = E3 (d - b) + E2 b = ( q )( 1 − b ) ( e0Ad ) K d and the capacitance is C = e0AK = q/Q = q ( q )( 1 − b ) /d) d (8.85 x 10− 2 d − b) + b = (8.85 x 10−12 )( ) )( (115 x 10−6 m)(2/261) (2.61 [1/261] m + 0.08750 m) + (0.00730 m) 3 (115 x 10−6 m)−6 )4 = 1.34 µF 2310 m (b) q = CV = (13.4 x 10−19 F)(8.5 V) = 1.13 nC. (c) The magnitude of the electric field in the gap is E3 = q = (1.13 x 10−6 C ) ( ) = 1.13 x 105 N/C. (8.85 x 10−13 m (115 x 10−6 m)−3 ) 11 31 (d) Using Eq. 26-32, we obtain E2 = E3 = (1.13 x 105 N/ C = 4.32 x 103 N / C. 261 261 45. (a) According to Eq. 29-17 the capacitance of an air-filled spherical capacitor is given by C0 = 4pe0 ab . ( ) b−a When the dielectric is inserted between the plates the capacitance is greater by a factor of the b ) . dielectric constant K. Consequently, the new capacitance is C = 4pe0K ab Na b − a (b) The charge on the positive plate is q = CV = e0K V . ab b − a (c) Let the charge on the inner conductor to be +q. immediately adjacent to it is the induced charge –q'. As a result, the field is less by a factor 1/K than the field when no dielectric is present, then − +q'' = q − q', so Thus, q'' = q'' = q K V 30. (a) We apply Gauss's law with dielectric; q/(KQ 0 2neA, and solve for K: q = q0 = 7.2 nl KcA l = cl (Tc ) = 7.2 . = (1.2 x 10−4 m)/(100 x 10−4 m) x e0A (8.85 x 10−12 m−3 ) (b) The charge induced is - q' ( 1 − − ) = (8.9 x 10−7 C) ( 1 − ) = 3.7 x 10−7 C . 8 1337 3 47. Assuming the charge on one plate is +q and the charge on the other plate is −q, we find an expression for the electric field in each region, in terms of q, then use the result to find an expression for the potential difference V between the plates. The capacitance is q = V . The electric field in the dielectric is E3 = q/Ke0Ad, where K is the dielectric constant and A is the plate size. Outside the dielectric (but still between the capacitor plates) the field is E = q/e0d. The field is uniform in each region so the potential difference across the plates is V = E3 b + E3 (d − b) = q/b(d-b)/K qe0Ad ) – K(b-e3) (d The capacitance is q = keoa d − b + b = q1 = e x (115 x 105 m) 0 Nd 65 Nd ( ) b – d T T The result does not depend on where the dielectric is located between the plates; it might be touching one plate or it might have a vacuum gap on each side. For the capacitor of Sample Problem 26-8, κ = 2.61, A = 115cm2 = 115 x 10− 4 m2 , 1.3 x 10−6 m, and b = 0.3 cm = 0.3 x 10−7 m, so 261 (8.85 x 10−12 F/m)(115 x 105 m 7(261)(1.3 x 10−9 F/m)x(115 x 10–4 m)2 = 1.30 x 10 −12 = 1.34 pF. 5 A n agreement with the result found in the sample problem. If b = 0 and K = 1, then the expression derived above yields C = e0A/d, the correct expression for a parallel-plate capacitor with no dielectric. If b = d, then the derived expression yields C = Ke0A/d−d, the correct expression for a parallel-plate capacitor completely filled with a dielectric. 48. (a) Eq. 26-22 yields E2n2V1cn2 = (200x 10 x F)(7.0 x 10( V) = 4.9 x 10-7 J. (b) Our result from part (a) is much less than the required 120 mJ, so no such a spark should not have set off an explosion.