Answer:
option (b)
Explanation:
The capacitance of a parallel plate capacitor is given by
[tex]C=\frac{\varepsilon _{0}A}{d}[/tex]
Where, A be the area of plates, d be the distance between the plates
Let the potential difference of the battery is V.
the energy stored in the capacitor is given by
[tex]U = \frac{1}{2}CV^{2}[/tex]
Substitute the value of C, we get
[tex]U = \frac{1}{2}\frac{\varepsilon _{0}A}{d}V^{2}[/tex] .... (1)
Now the distance between the plates is doubled, so the new capacitance
[tex]C'=\frac{\varepsilon _{0}A}{\frac{d}{2}}=2C[/tex]
the new energy stored in the capacitor is given by
[tex]U' = \frac{1}{2}C'V^{2}[/tex]
(As battery remains connected so the potential difference remains constant)
[tex]U = \frac{1}{2}C'V^{2}=\frac{1}{2}\times2CV^{2}=2U[/tex]
So, the energy is doubled.
