A capacitor has a circular plate of radiusr=12cm that are at a distance of .2cm part.this is hooked up to 12 V battery. if this caoacitor is submerged in water what happens to the capacitance?
C = e A/d
here A = pi r^2 = pi(.12)^2
and d = .002
where e is the permittivity of the material between the plates
in a vacuum or approximately in air e = ε0 = 8.8541878176.. × 10−12
however in water the e might be about 80 times as high and therefore the C would be 80 times as high.
That comes with the caveat that in the real world that water is likely to have contaminants in it that vastly reduce its resistivity so that despite the high capacitance in theory, current flows and it performs more as a resistor than a capacitor.