A) \[-2\sqrt{2}\]
B) - 1
C) 1
D) \[-\frac{1}{\sqrt{2}}\]
Correct Answer: A
Solution :
[a] \[\frac{KQ}{2{{a}^{2}}}=\sqrt{2}\frac{Kq}{{{a}^{2}}}\] |
\[\Rightarrow \]\[\left| \frac{Q}{q} \right|=2\sqrt{2}\] |
So, \[\frac{Q}{q}=-2\sqrt{2}\] |
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