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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 The capacitor is then disconnected from the charging circuit and the paper filling is withdrawn, allowing air to fill the gap between the plates.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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lublana

Answer:

[tex]V_1=9.25 V[/tex]

Explanation:

We are given that

[tex]V_0=2.5 V[/tex]

[tex]k=3.7[/tex]

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.

[tex]C_0=\frac{Q}{V_0}[/tex]

[tex]C'=kC=\frac{kQ}{V_1}[/tex]

[tex]\frac{Q}{V_0}=\frac{kQ}{V_1}[/tex]

[tex]V_0=\frac{V_1}{k}[/tex]

Using the formula

[tex]V_0=\frac{V_1}{k}[/tex]

[tex]V_1=V_0k=2.5\times 3.7=9.25 V[/tex]

Hence, the new potential difference [tex]V_1=9.25 V[/tex]

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