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A 3500 pF air-gap capacitor is connected to a 32 V battery.

If a piece of mica fills the space between the plates, how much charge will flow from the battery?

User Moo
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1 Answer

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The charge that will flow from the battery is \(1.12 \times 10^{-7} \, \text{C}\).

The charge (\(Q\)) on a capacitor is given by the formula:

\[ Q = C \cdot V \]

where:

\(Q\) = charge on the capacitor,

\(C\) = capacitance of the capacitor,

\(V\) = voltage across the capacitor.

Given:

\(C = 3500 \, \text{pF} = 3.5 \times 10^{-9} \, \text{F}\) (converting from picofarads to farads),

\(V = 32 \, \text{V}\).

\[ Q = (3.5 \times 10^{-9} \, \text{F}) \times (32 \, \text{V}) \]

\[ Q = 1.12 \times 10^{-7} \, \text{C} \]

Therefore, the charge that will flow from the battery is \(1.12 \times 10^{-7} \, \text{C}\).

User Hassan Siddiqui
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