A) \[1200N\]
B) \[400N\]
C) \[mg\,sin\,\theta \]
D) None of these
Correct Answer: A
Solution :
\[U=\frac{1}{2}kx_{0}^{2}=\frac{1}{2}k{{\left( \frac{qE}{k} \right)}^{2}}=\frac{{{q}^{2}}{{E}^{2}}}{2k}\]
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