Answer:
option C
Explanation:
Given,
Change on the capacitor = Q
Separation is doubled
Energy stored in the capacitor,E = ?
we know,
[tex]E = \dfrac{Q^2}{2C}[/tex]
and [tex]C = \dfrac{\epsilon_0A}{d}[/tex]
now,
[tex]E = \dfrac{Q^2}{2\epsilon A}\ d[/tex].......(1)
where d is the separation between the two plates.
now, when the separation is doubled
[tex]E' = \dfrac{Q^2}{2\epsilon A}\ (2d)[/tex]
[tex]E' = 2(\dfrac{Q^2}{2\epsilon_0 A}\ d)[/tex]
From equation (1)
E' = 2 E
Hence, the energy stored in the capacitor is doubled if the separation is increased.
The correct answer is option C.
