The plate area is doubled, and the plate separation is reduced to half its initial separation. What is the new charge on the negative plate

 Q = 4 Q₀

Explanation:

This is an exercise on capacitors, where the capacitance is

         C =  

if we apply the given conditions

         C = \epsilon_{o} \    \frac{2A}{0.5d}

         C = 4 \epsilon_{o} \    \frac{A}{d}

let's call the capacitance Co with the initial values

         C₀ = \epsilon_{o} \    \frac{A}{d}

         C = 4 C₀

The charge on each plate of a capacitor is

           Q = C ΔV

If the potential difference is maintained, the new charge is

       Q = 4 C₀ ΔV

let's call

        Q₀ = C₀ ΔV

we substitute

        Q = 4 Q₀

come closer and become far apart and become steam/gas