A parallel-plate capacitor is charged by a 9.00 V battery, then the battery is removed. Part A What is the potential difference between the plates after the battery is disconnected? Express your answer with the appropriate units. V V = nothing nothing SubmitRequest Answer Part B What is the potential difference between the plates after a sheet of Teflon is inserted between them? Express your answer with the appropriate units. V T V T = nothing nothing SubmitRequest Answer Provide Feedback Next

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aristocles aristocles · Answer 1 · 2019-09-27T02:31:57+02:00

Answer:

Part a)

V = 9 Volts

Part b)

[tex]\Delta V = 4.29 Volts[/tex]

Explanation:

Part a)

When capacitor plates are charged to a certain voltage then we have

[tex]\Delta V = 9 volts[/tex]

now battery is disconnected so here we have potential difference between the plates is given as

[tex]\Delta V = \frac{Q}{C}[/tex]

now we have charge conserved on the plates as well as the capacitance is also constant so here potential difference will remains the same

Part b)

Now the teflon sheet is inserted between the plates of capacitor

so capacitance of the capacitor will change by a factor of dielectric constant

So we will have

[tex]C' = kC[/tex]

[tex]k_{teflon} = 2.1[/tex]

so here we have new capacitance given as

[tex]C' = 2.1 C[/tex]

now the voltage difference will be given as

[tex]\Delta V = \frac{Q}{2.1C}[/tex]

[tex]\Delta V = \frac{9.00}{2.1}[/tex]

[tex]\Delta V = 4.29 Volts[/tex]