A certain substance has a dielectric constant of 2.8 and a dielectric strength of 18.0 MV/m. If it is used as the dielectric material in a parallel-plate capacitor, what minimum area should the plates of the capacitor have to obtain a capacitance of 7.00×10-8 F and to ensure that the capacitor will be able to withstand a potential difference of 3.5 kV?
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To solve this problem, we can use the formula for the capacitance of a parallel-plate capacitor:
C = (ε0 * εr * A) / d
Where:
C is the capacitance,
ε0 is the permittivity of free space (8.85 x 10^-12 F/m),
εr is the relative permittivity or dielectric constant of the substance,
A is the area of the plates, and
d is the distance between the plates.
We are given that C = 7.00 x 10^-8 F, εr = 2.8, and we need to find the minimum area (A) required. The distance between the plates (d) does not affect the area we are looking for, so we can ignore it for now.
Rearranging the formula, we can solve for A:
A = (C * d) / (ε0 * εr)
Now, let's calculate the required area:
Substituting the given values, we have:
A = (7.00 x 10^-8 F * d) / (8.85 x 10^-12 F/m * 2.8)
To ensure that the capacitor can withstand a potential difference (V) of 3.5 kV, we need to make sure the electric field strength between the plates does not exceed the dielectric strength (Es) of the substance.
The electric field strength (E) can be calculated using the formula:
E = V / d
Where:
E is the electric field strength,
V is the potential difference, and
d is the distance between the plates.
Substituting the values, we have:
E = (3.5 x 10^3 V) / d
Since the electric field strength (E) should not exceed the dielectric strength (Es = 18.0 x 10^6 V/m), we can set up the following inequality:
E ≤ Es
(3.5 x 10^3 V) / d ≤ 18.0 x 10^6 V/m
Simplifying, we have:
d ≥ (3.5 x 10^3 V) / (18.0 x 10^6 V/m)
Now, we have both the equation for the area and the inequality for the minimum distance. By combining these, we can find the minimum area required.
Let's calculate the minimum area:
Substituting the inequality value for d in the equation for A, we have:
A = (7.00 x 10^-8 F * (3.5 x 10^3 V) / (18.0 x 10^6 V/m)) / (8.85 x 10^-12 F/m * 2.8)
Simplifying, we find:
A = 0.0012475 m^2
Therefore, the minimum area that the plates of the capacitor should have to obtain a capacitance of 7.00 x 10^-8 F and ensure that the capacitor can withstand a potential difference of 3.5 kV is approximately 0.0012475 square meters.