Answer:
b. 1/2·C₀, 2·V₀
Explanation:
The capacitance on the parallel plate capacitor = C₀
The area of the plates = A
The voltage on the battery = V₀
The magnitude of the charge on the plate = Q₀
The new distance between the plates = 2·d
From an online physics source, we have;

Where;
ε₀ = Constant
A = The area of the plates
With the new distance, 2·d, we get;

Therefore;

The potential difference, 'V', is given as follows;

Therefore;

Given that Q = Q₀, we get;

∴ V = 2 × V₀
The new potential difference, V = 2·V₀
Therefore, after the plates are 2·d apart, the new capacitance and potential difference between the plates are;
1/2·C₀, 2·V₀.