Answer :
Final answer:
The equipotential surface at a voltage of 6.00V would be located halfway between the two plates of the parallel-plate capacitor, about 3.00 cm away from the negative plate.
Explanation:
In a parallel-plate capacitor, the voltage is distributed evenly across the distance between the two plates. So if you have a 12.0V potential difference across the plates and they are 6.00 cm apart, each cm corresponds to a 2.00V potential difference (12.0V / 6.00 cm = 2.00V/cm).
The equipotential surface at a potential of 6.00V relative to the negative plate would then be located 3.00 cm away from the negative plate (6.00V / 2.00V/cm = 3.00cm).
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