<p>快速且“肮脏”的修复:
您可以将代码中的“use”替换为“k1=”。。。
这将把k1值设置为您期望的值,代码将正常工作</p>
<p>正如其他评论所建议的那样,您仍然可以改进编码风格</p>
<pre><code>A1= float(input("What is the value of the area of the plates used in the first capacitor ? \n"))
if A1 == 0:
print("The area cannot be zero.")
raise SystemExit
d1= float(input("What is the separation between the plates used in the first capacitor ? \n"))
if d1 == 0:
print("The distance cannot be zero.")
raise SystemExit
k1= input ('What is the medium used between the plates of first capacitor (vacuum, Air, polystyrene, paper, silicon, pyrex glass, porcelain, nerve membrane, ethanol, water \n)')
if k1== 'vacuum':
print (1)
elif k1 == 'Air':
k1=(1.0005)
elif k1 == 'polystyrene':
k1=(2.6)
elif k1 == 'paper':
k1=(3.5)
elif k1 == 'silicon':
k1=(12.0)
elif k1 == 'pyrex glass':
k1=(4.7)
elif k1 == 'porcelain':
k1=(6.5)
elif k1 == ' nerve membrane':
k1=(7.0)
elif k1 == 'ethanol':
k1=(25.0)
elif k1 == 'water':
k1=(78.5)
E0= 8.85*10**-12
C1= k1* E0*A1/d1
print ('E0=', E0, 'C1=', C1)
</code></pre>
<p>它起作用了,结果如下:</p>
<pre><code>What is the value of the area of the plates used in the first capacitor ?
5
What is the separation between the plates used in the first capacitor ?
2
What is the medium used between the plates of first capacitor (vacuum, Air, polystyrene, paper, silicon, pyrex glass, porcelain, nerve membrane, ethanol, water
)water
E0= 8.849999999999999e-12 C1= 1.7368124999999997e-09
</code></pre>
<p>对于另一个错误,我可以在这里向您展示缩进的正确方式,
但是我没有您的代码的剩余部分,并且在注释中很难阅读:因此,这里是一个尝试:</p>
<p>缩进应为4个空格,如下所示:</p>
<pre><code>if Ceq == 'series':
Ceq = (C1**-1 + C2**-1)**-1
elif Ceq == 'parallel':
Ceq= C1+ C2
</code></pre>