Two parallel plates are separated by a dielectric of thickness 2 mm with a dielectric constant of 6. If the area of each plate is 40 cm^2 and the potential difference between them is 1.5 kV, determine (a) the capacitance
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To determine the capacitance (C) of the parallel plate capacitor, we can use the formula:
C = (ε₀ * εᵣ * A) / d
where:
C = capacitance
ε₀ = permittivity of free space (8.85 x 10^-12 F/m)
εᵣ = relative permittivity or dielectric constant
A = area of each plate
d = separation between the plates
In this case, we have the following values:
ε₀ = 8.85 x 10^-12 F/m
εᵣ = 6
A = 40 cm^2 = 40 x 10^-4 m^2 (convert cm^2 to m^2)
d = 2 mm = 2 x 10^-3 m (convert mm to m)
Now we can substitute these values into the formula to calculate the capacitance:
C = (8.85 x 10^-12 F/m * 6 * 40 x 10^-4 m^2) / 2 x 10^-3 m
Simplifying the equation:
C = (8.85 x 6 x 40 x 10^-12 x 10^-4) / 2 x 10^-3
Next, multiply the values inside the parentheses and simplify the exponents:
C = (8.85 x 6 x 40 x 10^-16) / 2 x 10^-3
C = (212.4 x 10^-16) / 2 x 10^-3
Finally, simplify the expression:
C = 106.2 x 10^-13 / 2 x 10^-3
C = 53.1 x 10^-13 / 10^-3
C = 53.1 x 10^-10 F
Thus, the capacitance of the parallel plate capacitor is 53.1 pF (picoFarads).