Question

我正在为我的物理2类做一个Vpython分配，要求对偶极子的电场进行编程。我已经编写了以下代码：

## constants
oofpez = 9e9 # stands for One Over Four Pi Epsilon-Zero
qe = 1.6e-19  # postive charge value
s = 4e-11    # charge separation
R = 3e-10  # display Enet on a circle of radius R
scalefactor = 3e-20 # for scaling arrows to represent electric field

## objects
## Represent the two charges of the dipole by red and blue spheres:
plus = sphere(pos=vector(s/2,0,0), radius=1e-11, color=color.red)
qplus = qe   # charge of positive particle
neg = sphere(pos=vector(-s/2,0,0), radius=1e-11, color=color.blue)
qneg = -qplus  # charge of negative particle


## calculations
## You will complete the lines required to make a loop calculate and display the net dipole electric field
## at equally spaced angles on a circle radius R around the dipole. The dipole is centered at the origin.
theta = 0
while theta < 2*pi:
    rate(2)   # tell computer to go through loop slowly
    ## Calculate observation location (tail of arrow) using current value of theta:
    Earrow = arrow(pos=R*vector(cos(theta),sin(theta),0), axis=vector(1e-10,0,0), color=color.orange)
    ## assign the name TestLocation to be the observation location on the circle radius R
    TestLocation=R*vector(cos(theta),sin(theta),0)
    ## write instructions below to tell the computer how to calculate the correct 
    ## net electric field Enet at the observation location (the position of Earrow):
    rPlus=TestLocation-plus.pos
    rPlusMag=((R*cos(theta)-(s/2))^2+(R*sin(theta))^2)^0.5
    rPlusHat=rPlus/rPlusMag
    Eplus=oofpez*qplus/(rPlusMag)^2*rPlusHat

    rNeg=TestLocation-neg.pos
    rNegMag=((R*cos(theta)-(-s/2))^2+(R*sin(theta))^2)^0.5
    rNegHat=rNeg/rNegMag
    Eneg=oofpez*qneg/(rNegMag)^2*rNegHat

    Etotal=Eplus+Eneg

    Enet=arrow(pos=TestLocation,axis=Etotal*scalefactor, color=color.green)

    ## change the axis of Earrow to point in the direction of the electric field at that location
    ## and scale it so it looks reasonable
    ## Efield = arrow(pos=R*vector(cos(theta),sin(theta),0), axis=Etotal*scalefactor, color=color.blue)
    Earrow.axis=Etotal*scalefactor

    ## Assign a new value to theta
    theta = theta + pi/6

该分配具有一个预制的模板，该模板使用注释创建，声明了适当的变量，并且所述变量分配了正确的值，因此，从理论上讲，如果我正确输入其余代码，它应该可以正确运行。我编写的代码以“ rPlus = ...”开头，以“ Enet = ...”结尾，但是，当我运行它（使用GlowScript IDE）时，它给出一条错误消息，提示“错误：属性'axis'必须是向量”，我确信这意味着在代码的该部分中分配给Enet.axis的值有问题。我浏览了我生成的代码，但似乎找不到错误。

我们正在学习python作为常规课程的补充，因此除这些作业外，我没有python的背景知识。我不需要寻找电场的帮助，而是为什么会弹出错误消息。在正确方向上的任何帮助或提示，将不胜感激！

谢谢

Answer 1

在第31行，您使用^而不是**进行求幂。

如果您使用的是Chrome浏览器，则可以获得错误的行号。

我通过在代码的关键位置插入打印语句发现了问题，这发现rPlusMag为0，而不是向量。

Vpython：属性“ axis”必须是向量

1 个答案: