关于有一道物理电磁场的问题（英文题目）An uncharged conducting cylinder of infinite length and radius a is placed in a previously uniform electric field E in vacuo with its axis at right angles to the eletric vector.show that v=Arco
来源：学生作业帮助网 编辑：六六作业网 时间：2024/11/22 10:20:09
关于有一道物理电磁场的问题（英文题目）An uncharged conducting cylinder of infinite length and radius a is placed in a previously uniform electric field E in vacuo with its axis at right angles to the eletric vector.show that v=Arco
关于有一道物理电磁场的问题（英文题目）
An uncharged conducting cylinder of infinite length and radius a is placed in a previously uniform electric field E in vacuo with its axis at right angles to the eletric vector.show that v=Arcosx+(B/r)cosx is a possible distribution of the electrosatic potential in appropriate cylinderical co-ordinates for r>a,and obtain the values of the constants A and B.
关于有一道物理电磁场的问题（英文题目）An uncharged conducting cylinder of infinite length and radius a is placed in a previously uniform electric field E in vacuo with its axis at right angles to the eletric vector.show that v=Arco
siminghao的翻译是准确的
你把无限长的电线放到了匀强电场里,假设这条电线是竖着放的.很明显你任取一条平行于电线的线,这条线上的电势都是一样的,所以电势和高度没关系.所以可以断定题目中给的柱坐标的圆柱轴线是平行于电线的.
导体放电场里,导体表面等势,导体表面场线垂直表面.
把这个东西换成直角坐标系.x=rcosθ,y=rsinθ,v=Ax+Bx/(x^2+y^2).导线不带电,所以电势是0,方程0=Ax+Bx/(x^2+y^2),正好是个圆心在原点的圆x^2+y^2=-B/A,所以a^2=-B/A.当r->无穷,dv/dr的最大值就是场强,所以E=A.相应B=-E*a*a.
一个无限长半径为a的不带电的圆柱形导体，置于真空中均匀电场E中，圆柱体的轴与电场矢量方向垂直，用柱坐标证明当r>a时v=Arcosx+（B/r)cosx 是电势能的一个可能的分布,并求出常量A和B的值。
会翻译，不会做！
一个不带电的无限长度进行缸和半径的一个被置于均匀电场的先前与地轴vacuo成直角的电气vector.表明v = Arcosx +(B)cosx可能分布在适当cylinderical electrosatic潜力为>是统筹,并获得了价值观的常量,a和B。
SAT题？