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Anna [14]
3 years ago
11

Memories

Physics
1 answer:
nevsk [136]3 years ago
7 0

Answer:

Its ok but the answer is C

Explanation:

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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
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Answer:

Q=3.9825\times 10^{-9} C

Explanation:

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

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d=Initial distance between plates

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Potential difference between plates after moving

V=\frac{Q}{C_1}

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\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

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

\frac{Q\Delta d}{\epsilon_0 S}=100

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Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

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

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