A certain dielectric, with a dielectric constant, 25, can withstand an electric field of 7×10^7 V/m. Suppose we want to use this dielectric to construct a 0.3 microFarad capacitor that can withstand a potential difference of 2000 V.

The permittivity of free space is 8.85×10^-12 C^2/N*m^2.

What is the minimum plate separation?
What must the area of the plates be? Answer in units m^2







The full question states, A parallel plate capacitor is connected to a battery. If we double the plate separation,

1. the capacitance is doubled
2. None of these
3. the charge on each plate is halved
4. the potential difference is halved
5. the Electric field is doubled