A parallel-plate capacitor is made of two flat metal plates pressed against a thin slab of dielectric material. The capacitor is connected t

Question

A parallel-plate capacitor is made of two flat metal plates pressed against a thin slab of dielectric material. The capacitor is connected to a power supply, and a potential difference of 90 V is applied to the plates. With the power supply disconnected, the dielectric material is removed and the potential difference between the plates is measured to be 525 V. What is the dielectric constant of the material that was initially used to fill the gap between the plates?

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Delwyn 3 years 2021-08-24T22:31:52+00:00 1 Answers 25 views 0

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  1. nguyetanh
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    2021-08-24T22:33:22+00:00
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    Answer:

    5.8

    Explanation:

    • The capacitance of any capacitor, by definition, is as follows:

            C =\frac{Q}{V}

    • For a parallel-plate capacitor, applying Gauss’Law and the definition of  electric potential, it can be found that the capacitance can be expressed as follows:

           C =\frac{\epsilon*A}{d}

    • Where ε = ε₀*εr (being εr the dielectric constant of the material that fills the space between the plates, A the area of one of the plates, and d, the separation between them).
    • If the capacitor is disconnected from the battery, due to the conservation of the charge, the charge can’t change after being disconnected, so Qf = Q₀.
    • If we apply the definition of capacitance, we can find the relationship between the initial and the final capacitance, as follows:

            C_{0} = \frac{Q_0}{V_{0} } \\ \\ C_{f} = \frac{Q_f}{V_{f} } \\ \\ Q_{0} = Q_{f} = C_{0} * V_{0} = C_{f} * V_{f} \\ \\ \frac{C_{0} }{C_{f} } = \frac{V_{f}}{V_{0}} = \frac{525 V}{90V} = 5.8

    • If we look at the expression for the capacitance in a parallel-plate capacitor, if  A and d remain constant, the only parameter changing is the dielectric constant ε, so we can write the following  equation:

            \frac{C_{0}}{C_{f} } = \frac{\epsilon r*\epsilon 0*A}{d} *\frac{d}{\epsilon 0*A} =5.8

            ⇒εr = 5.8

    • The dielectric constant of the material that was initially used to fill the gap between the plates is 5.8.

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