A paper-filled capacitor is charged to a potential difference of V 0 = 2.5 V V0=2.5 V . The dielectric constant of paper is κ = 3.7 κ=3.7 .

Question

A paper-filled capacitor is charged to a potential difference of V 0 = 2.5 V V0=2.5 V . The dielectric constant of paper is κ = 3.7 κ=3.7 . The capacitor is then disconnected from the charging circuit and the paper filling is withdrawn, allowing air to fill the gap between the plates. Find the new potential difference V 1 V1 of the capacitor.

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niczorrrr 3 years 2021-09-05T13:04:25+00:00 1 Answers 42 views 0

Answers ( )

  1. Cherry
    0
    2021-09-05T13:05:33+00:00
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    Answer:

    V_1=9.25 V

    Explanation:

    We are given that

    V_0=2.5 V

    k=3.7

    We have to find the new potential difference of the capacitor.

    When the capacitor is disconnected then the charge stored in capacitor is constant.

    When we introduce material of dielectric constant k between the plates of capacitor then the capacitance of capacitor increases k times.

    C_0=\frac{Q}{V_0}

    C'=kC=\frac{kQ}{V_1}

    \frac{Q}{V_0}=\frac{kQ}{V_1}

    V_0=\frac{V_1}{k}

    Using the formula

    V_0=\frac{V_1}{k}

    V_1=V_0k=2.5\times 3.7=9.25 V

    Hence, the new potential difference V_1=9.25 V

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