A) \[\frac{U}{2}\]
B) \[U\]
C) \[2\,\,U\]
D) \[4\,\,U\]
Correct Answer: C
Solution :
Energy stored \[U=\frac{1}{2}qV\] As the distance d is increased between the two plates. Now, stored energy, \[U'=\frac{1}{2}qV'\] \[=\frac{1}{2}q\left[ \frac{q}{C} \right]\] \[=\frac{1}{2}\frac{{{q}^{2}}d}{{{\varepsilon }_{0}}A}\] \[U\propto d\] Hence,- \[U'=2U\]
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