A) \[0.1\,m\]
B) \[0.04\,m\]
C) \[0.033\,m\]
D) \[0.33\,m\]
Correct Answer: C
Solution :
Let neutral point be obtained at a distance x from 20 mC charge. Hence at neutral point \[\frac{20}{{{(x)}^{2}}}=\frac{80}{{{(10-x)}^{2}}}\] Þ \[x=+\,0.033\,m\]You need to login to perform this action.
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