A parallel plate air-filled capacitor has a capacitance of 2 μF . When it is half-filled with a dielectric of dielectric constant k = 3, its capacitance becomes
Question
A parallel plate air-filled capacitor has a capacitance of 2 μF . When it is half-filled with a dielectric of dielectric constant k = 3, its capacitance becomes
Solution
The capacitance of a parallel plate capacitor is given by the formula:
C = ε0 * (A/d)
where: C is the capacitance, ε0 is the permittivity of free space, A is the area of one of the plates, and d is the distance between the plates.
When a dielectric is introduced, the capacitance increases by a factor of the dielectric constant (k). So if the capacitor is half-filled with a dielectric, the total capacitance is the average of the capacitance with and without the dielectric.
So, the new capacitance C' is given by:
C' = 1/2 * (C + k*C)
Substituting the given values:
C' = 1/2 * (2 μF + 3 * 2 μF) = 1/2 * (2 μF + 6 μF) = 4 μF
So, when the capacitor is half-filled with a dielectric of dielectric constant k = 3, its capacitance becomes 4 μF.
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