《大学物理课后习题答案14电磁场习题.ppt

* 电 磁 场 习 题 14-1 14-2 14-3 14-4 14-5 14-6 14-7 14-8 14-9 14-10 习题总目录 结束 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 14-1 试证明平行板电容器中的位移电流 可写为： 式中C是电容器的电容，U是两极板间的电势 差。如果不是平行板电容器，上 式可以应用 吗? 如果是圆柱形电容器，其中的位移电流 密度和平板电容器时有何不同? = C I t d d U d 目录 结束 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 证：设极板面积S，板间距d 若不是平行板电容器，上式仍可适用。 位移电流密度 平行板电容器 证： = U Id t C d d Φ S d = U = C d d t U D δ d d d t = σ = D d t = δ d σ d C U Φ = e 0 d d = Φ C t e 0 ∴ 圆柱形电容器 l = D 2 r p 1 2 r d t = δ d d l p 目录 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 14-2 在一对巨大的圆形极板(电容C=l.0 ×10-12 F)上，加上频率为50Hz、峰值为 174000V的交变电压，计算极板间位移电 流的最大。 目录 结束 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 已知：C =1.0×10-12 F, f =50Hz , Um =1.74×105V 求：Idm 解： E S = Φ C = d S e 0 ω cos = d t S Um Φ t d = d e 0 Id sin C = t ω ω Um Id ∴ Id 的最大值 =1.0×10-12×2p×50×1.74×105 =5.74×10-5(A) C = ω Idm Um 2 f = C p Um 目录 结束 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 14-3 有一电荷q ，以速度v(vc)作匀速 运动。试从 . ò Ψ = H t l d d d 计算离电荷r 处的磁场强度。 目录 结束 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 解：通过平面的电位移通量 4 π r 2 q 2 a q sin = 0 d ò π r 2 q D S d . = Φ s ò ò = d S 4 π r 2 q s ò ò 2 cos q = ( ) 1 a 已知： q , v (vc) 从 求： ò H l . d = Φ t 0 d d H a q v R 目录 结束 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 电荷在运动，a 在变化 d q = sin 2 d t a a q v = sin 2 a r 2 由于对称性在半径为R 的平面上H值相同 v = sin a d t d a r ∵ I = Φ t 0 d d d ∴ q . ò d = H sin 2 a d t d a l ∴ 目录 结束 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. C