Final answer:
To find the minimum plate area of a capacitor with a dielectric constant of 2.8 to achieve a capacitance of 9.00×10−8 F and withstand a potential difference of 5.3 kV, one must calculate the plate separation based on dielectric strength and use the capacitance formula, resulting in an area of approximately 0.00106 m².
Step-by-step explanation:
To determine the minimum area of the plates for the capacitor, we use the formula for capacitance C for a parallel-plate capacitor with a dielectric:
C = ε0 κ A / d
Where:
- C is the capacitance,
- ε0 is the permittivity of free space (8.85×10−12 F/m),
- κ is the dielectric constant,
- A is the area of the plates, and
- d is the separation between the plates.
Since the dielectric strength is 18.0 MV/m and the capacitor must withstand a potential difference of 5.3kV, the separation d can be found using:
V = E × d
Solving for d, we have:
d = V / E = 5.3×103 V / 18.0×106 V/m = 2.94×10−5 m
Now we can substitute C, ε0, κ, and d into the capacitance formula and solve for A:
A = C × d / (ε0 κ) = (9.00×10−8 F) × (2.94×10−5 m) / (8.85×10−12 F/m × 2.8) ≈ 0.00106 m2
Therefore, the minimum area of the plates must be approximately 0.00106 m2, or 1060 mm2, to achieve the desired capacitance and withstand the potential difference without breakdown.