A parallel-plate capacitor is constructed with circular plates of radius 5.6×10−2 m . The plates are separated by 0.27 mm , and the space between the plates is filled with a dielectric with dielectric constant κ. When the charge on the capacitor is 1.9 μC the potential difference between the plates is 780 V . Find the value of the dielectric constant, k.
Isn't this a standard formula?
C=k*episilon*Area/separation
C=Q/V set them equal, solve for k
To find the value of the dielectric constant (κ), we can use the equation that relates the capacitance (C) of a parallel-plate capacitor to its geometric properties:
C = (κ * ε₀ * A) / d
Where:
C is the capacitance of the capacitor,
κ is the dielectric constant,
ε₀ is the permittivity of free space (a constant equal to 8.85 x 10^-12 F/m),
A is the area of the plates,
and d is the separation between the plates.
We are given the following values:
Charge, q = 1.9 μC = 1.9 x 10^-6 C
Potential difference, V = 780 V
Radius, r = 5.6 x 10^-2 m
Separation, d = 0.27 mm = 0.27 x 10^-3 m.
The area of a circular plate is given by:
A = πr²
Plugging in the given values, we can calculate the area:
A = π(5.6 x 10^-2)^2
Next, we can use the equation relating capacitance, charge, and potential difference:
C = q / V
Plugging in the given values, we can calculate the capacitance:
C = (1.9 x 10^-6) / 780
Now we can rearrange the formula for capacitance to solve for the dielectric constant:
κ = (C * d) / (ε₀ * A)
Plugging in the known values, we can calculate the dielectric constant κ.