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A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the voltage source. When the plates are moved 0.8 cm farther apart, the voltage between the plates increases by 100 V. What is the charge Q on the positive plate of the capacitor?

Answer :

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Answer:

[tex]Q=3.9825\times 10^{-9} C[/tex]

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=[tex]\frac{360}{10000}=0.036 m^2[/tex]

[tex]\Delta d=0.8 cm=0.008 m[/tex]

[tex]\Delta V=100 V[/tex]

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

[tex]C=\frac{\epsilon_0 S}{d}[/tex]

Capacitance of capacitor after moving plates

[tex]C_1=\frac{\epsilon_0 S}{(d+\Delta d)}[/tex]

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

Potential difference between plates after moving

[tex]V=\frac{Q}{C_1}[/tex]

[tex]V+\Delta V=\frac{Q}{C_1}[/tex]

[tex]\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}[/tex]

[tex]\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100[/tex]

[tex]\frac{Q\Delta d}{\epsilon_0 S}=100[/tex]

[tex]\epsilon_0=8.85\times 10^{-12}[/tex]

[tex]Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}[/tex]

[tex]Q=3.9825\times 10^{-9} C[/tex]

Hence, the charge on positive plate of capacitor=[tex]Q=3.9825\times 10^{-9} C[/tex]

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