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内外半径分别为a和b的无限长圆柱形电容器单位长度荷电为λf 板间填充导率为σ的非磁性物质。（1) 证明在介质

大学 2022-05-21 03:32:27 圆柱形,介质,电容器已帮助人

内外半径分别为a和b的无限长圆柱形电容器，单位长度荷电为λ_f，板间填充导率为σ的非磁性物质。

(1) 证明在介质中任何一点传导电流与位移电流严格抵消，因此内部无磁场。

(2) 求λ_f随时间的衰减规律。

(3) 求与轴相距为r的地方的能量耗散功率密度。

(4) 求长度为ι的一段介质总的能量耗散功率，并证明它等于这段的静电能减少率。

热搜：本票和支票的付款人分别为( )

答案解析

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电荷分布呈轴对称，且自由电荷只分布在导体面。
以圆柱中心为z轴，利用高斯定理，得介质内任一点。
D满足
故 $latex.action?latex=RD1cZnJhY3tcbGFtYmRhIF97Zn1yfXsyXHBpIHJeezJ9fQ%3D%3D$ (r为z轴指向)
各向同性线性介质满足 D=εE
故 $latex.action?latex=RT1cZnJhY3tcbGFtYmRhIF97Zn19ezJccGkgXHZhcmVwc2lsb24gfVxmcmFje3J9e3JeezJ9fQ%3D%3D$
由欧姆定律传导电流 $latex.action?latex=Sl97Zn09XHNpZ21hIEU9XGZyYWN7XHNpZ21hIFxsYW1iZGEgX3tmfX17MlxwaSBcdmFyZXBzaWxv%0D%0AbiB9XGZyYWN7cn17cl57Mn19$
而位移电流 $latex.action?latex=Sl97RH09XGZyYWN7XHBhcnRpYWwgRH17XHBhcnRpYWwgdH09XGZyYWN7MX17MlxwaX1cZnJhY3ty%0D%0AfXtyXnsyfX1cZnJhY3tccGFydGlhbCB9e1xwYXJ0aWFsIHR9IFxsYW1iZGEgX3tmfQ%3D%3D$
因 $latex.action?latex=XGJpZ3RyaWFuZ2xlZG93biA%3D$ ·D=ρ由电荷守恒定律可知
$latex.action?latex=XGZyYWN7XHBhcnRpYWwgXHJobyB9e1xwYXJ0aWFsIHR9K1xiaWd0cmlhbmdsZWRvd24gXGNkb3Qg%0D%0ASj0w$
而 $latex.action?latex=XHJobyA9XGJpZ3RyaWFuZ2xlZG93biBcY2RvdCBEPVxmcmFje1xsYW1iZGEgX3tmfX17MlxwaSB9%0D%0AXGJpZ3RyaWFuZ2xlZG93biBcY2RvdCBcZnJhY3tyfXtyXnsyfX09XGZyYWN7XGxhbWJkYSBfe2Z9%0D%0AfXsyXHBpIFx2YXJlcHNpbG9uIHJeezJ9fQ%3D%3D$
$latex.action?latex=XGJpZ3RyaWFuZ2xlZG93biBcY2RvdCBKX3tmfT1cZnJhY3tcc2lnbWEgX3tmfX17MlxwaSBcdmFy%0D%0AZXBzaWxvbiB9XGJpZ3RyaWFuZ2xlZG93biBcY2RvdCBcZnJhY3tyfXtyXnsyfX09XGZyYWN7XHNp%0D%0AZ21hIFxsYW1iZGEgX3tmfX17MlxwaSBcdmFyZXBzaWxvbiByXnsyfX0%3D$
所以 $latex.action?latex=XGZyYWN7MX17MlxwaSByXnsyfX1cbGVmdCggXGZyYWN7XHBhcnRpYWwgXGxhbWJkYSBfe2Z9fXtc%0D%0AcGFydGlhbCB0fSArXGZyYWN7XHNpZ21hIFxsYW1iZGEgX3tmfX17XHZhcmVwc2lsb24gfVxyaWdo%0D%0AdCk9MA%3D%3D$
$latex.action?latex=XGZyYWN7XHBhcnRpYWwgXGxhbWJkYSBfe2Z9fXtccGFydGlhbCB0fT1cZnJhY3tcc2lnbWEgXGxh%0D%0AbWJkYSBfe2Z9fXtcdmFyZXBzaWxvbiB9$
$latex.action?latex=Sl97RH0rSl97Zn09XGZyYWN7MX17MlxwaSB9XGZyYWN7cn17cl57Mn19XGxlZnQoIFxmcmFje1xw%0D%0AYXJ0aWFsIH17XHBhcnRpYWwgdH1cbGFtYmRhIF97Zn0gK1xmcmFje1xzaWdtYSB9e1x2YXJlcHNp%0D%0AbG9uIH1cbGFtYmRhIF97Zn1ccmlnaHQpPTA%3D$
即传导电流与位移电流相抵消。
(2) (1) 中已得到 $latex.action?latex=XGZyYWN7XHBhcnRpYWwgfXtccGFydGlhbCB0fVxsYW1iZGEgX3tmfStcZnJhY3tcc2lnbWEgfXtc%0D%0AdmFyZXBzaWxvbiB9XGxhbWJkYSBfe2Z9PTA%3D$
解微分方程得 $latex.action?latex=XGxhbWJkYSBfe2Z9PVxsYW1iZGEgX3tmX3swfX1lXnstXGZyYWN7XHNpZ21hIH17XHZhcmVwc2ls%0D%0Ab24gfXR9$
λf0为t=0时刻，λf的值。
(3) 介质内的能量耗散是由电流的热效应引起的，故耗散能量密度为
$latex.action?latex=Sl97Zn1cY2RvdCBFPVxzaWdtYSBFXnsyfT1cc2lnbWEgXGxlZnQoIFxmcmFje1xsYW1iZGEgX3tm%0D%0AfX17MlxwaSBcdmFyZXBzaWxvbiByfSBccmlnaHQpXnsyfQ%3D%3D$
(4)长度为，的介质内总的能耗功率为
$latex.action?latex=UD1caW50IFxzaWdtYSBFXnsyfWRcdGF1PVxpbnRfe2F9XntifVxzaWdtYSBFXnsyfSgyXHBpIFxp%0D%0Ab3RhIHIpZHI9XGludF97YX1ee2J9XGZyYWN7XHNpZ21hIFxsYW1iZGEgX3tmfV57Mn1caW90YSB9%0D%0AezJccGkgXHZhcmVwc2lsb24gXnsyfX1cZnJhY3tkcn17cn09XGZyYWN7XGlvdGEgXHNpZ21hIFxs%0D%0AYW1iZGEgX3tmfV57Mn19ezJccGkgXHZhcmVwc2lsb24gXnsyfX1caW50X3thfV57Yn1cZnJhY3tk%0D%0Acn17cn09XGZyYWN7XGlvdGEgXHNpZ21hIFxsYW1iZGEgX3tmfV57Mn19ezJccGkgXHZhcmVwc2ls%0D%0Ab24gXnsyfX1sblxmcmFje2J9e2F9$
场的能量密度 $latex.action?latex=Vz1cZnJhY3sxfXsyfShFXGNkb3QgRCtIXGNkb3QgQik%3D$
(1)中已证得介质内无磁场
故 $latex.action?latex=Vz1cZnJhY3sxfXsyfShcdmFyZXBzaWxvbiBFXnsyfSk9XGZyYWN7XGxhbWJkYSBfe2Z9XnsyfX17%0D%0AOFxwaSBeezJ9XHZhcmVwc2lsb24gfVxmcmFjezF9e3JeezJ9fQ%3D%3D$
能量减少率为 $latex.action?latex=LVxpbnQgXGZyYWN7XHBhcnRpYWwgXG9tZWdhIH17XHBhcnRpYWwgdH1kXHRhdSA9LVxpbnRfe2F9%0D%0AXntifVxmcmFje1xsYW1iZGEgX3tmfX17NFxwaSBeezJ9XHZhcmVwc2lsb24gcl57Mn19XGNkb3Qg%0D%0AXGZyYWN7XHBhcnRpYWwgXGxhbWJkYSBfe2Z9fXtccGFydGlhbCB0fVxjZG90IDJccGkgcmUgZHI9%0D%0AXGZyYWN7XHNpZ21hIFxpb3RhIFxsYW1iZGEgX3tmfV57Mn19ezJccGkgXHZhcmVwc2lsb24gXnsy%0D%0AfX1caW50X3thfV57Yn1cZnJhY3tkcn17cn09XGZyYWN7XHNpZ21hIFxpb3RhIFxsYW1iZGEgX3tm%0D%0AfV57Mn19ezJccGkgXHZhcmVwc2lsb24gXnsyfX1sblxmcmFje2J9e2F9$
由此可见介质内总的能量耗散功率等于静电能减少率。