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in Electric Potential and Capacitance by (58.8k points)
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A parallel plate capacitor with air between the plates has a capacitance of 8pF (1 pF=10-12F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant s?

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The capacitance of capacitor with air as dielectric is given by

C = \(\frac{\varepsilon_0A}{d}\)

Given C = 8pF = 8 × 10-12F……(1)

If C is new capacitance when d1 = \(\frac{d}{2}\)and space is filled with a substance of dielectric constant k = 6. Then

C1 = \(\frac{\varepsilon_0kA}{d_1}=\frac{\varepsilon_0kA}{d/2}\)

or C1 = \(\frac{2k\varepsilon_0A}{d}\)

Using Eq.(1)

C1 = 12 × 8 × 10-12 F

or C1 = 96pF.

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