作业帮关于有一道物理电磁场的问题（英文题目）
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/14 20:33:45
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.
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.
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.
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.