A parallel-plate capacitor is constructed with circular plates of radius 0.056 m. The plates are separated by 0.25 mm, and the space between the plates is filled with dielectric with dielectric constant k. When the charge on the capacitor is 1.2 uc the potential difference between the plates is 750 V. Find the value of the dielectric constant, k.
To find the value of the dielectric constant, k, we can make use of the formula for capacitance in a parallel-plate capacitor:
C = (k * ε₀ * A)/d
where C is the capacitance, k is the dielectric constant, ε₀ is the permittivity of free space (8.85 * 10^(-12) F/m), A is the area of the plates, and d is the separation between the plates.
First, we need to find the capacitance of the capacitor. We can rearrange the formula above to solve for capacitance:
C = (Q / V)
where Q is the charge on the capacitor and V is the potential difference between the plates.
Substituting the given values:
C = (1.2 * 10^(-6) C) / (750 V)
C ≈ 1.6 * 10^(-9) F
Now, let's calculate the area of the plates using the formula for the area of a circle:
A = π * r^2
where r is the radius of the circular plates given as 0.056 m.
A = π * (0.056 m)^2
A ≈ 9.862 * 10^(-3) m^2
Now, substitute the values into the formula for capacitance:
1.6 * 10^(-9) F = (k * (8.85 * 10^(-12) F/m) * (9.862 * 10^(-3) m^2)) / (0.25 * 10^(-3) m)
Solving for k:
k = (1.6 * 10^(-9) F * (0.25 * 10^(-3) m)) / ((8.85 * 10^(-12) F/m) * (9.862 * 10^(-3) m^2))
k ≈ 18.1
Therefore, the value of the dielectric constant, k, is approximately 18.1.