Answer:
Potential difference between the plates will increases.
Explanation:
Given that ,first battery charged the capacitor then is removed.
We know that
Capacitance C
[tex]C=\dfrac{A}{4\pi \varepsilon d}[/tex]
[tex]C\alpha \dfrac{1}{ d}[/tex]
And we also know that
Q=C ΔV
ΔV=Q/C
So when distance between plates is increase then it will reduce capacitance(C).It means that for maintaining constant charge on capacitor potential difference will be increase.