The capacity between two adjacent plates of parallel plate condenser is 5 - F- The number of plates used if we want to get a capacity of 40 - F are alternate plates are connected together-

The correct option is C 9 Connecting n plates alternatively creates a capacitance of (n 1)Aϵ_0d .Given, capacity of each condenser: ϵ_0 Ad ...

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Standard XII

Physics

Capacitance of Parallel Plate Capacitor

The capacity ...

Question

The capacity between two adjacent plates of parallel plate condenser is $5 μ F$ . The number of plates used if we want to get a capacity of $40 μ F$ are (alternate plates are connected together):

8

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9

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7

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Solution

The correct option is C $9$
Connecting $n$ plates alternatively creates a capacitance of $\frac{(n - 1) A ϵ_{0}}{d}$ .
Given, capacity of each condenser: $\frac{ϵ_{0} A}{d} = 5 μ F$
So,
$\frac{(n - 1) A ϵ_{0}}{d} = 40 μ F$
$\Rightarrow n = 9$

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The capacity between two adjacent plates of parallel plate condenser is 5 μF. The number of plates used if we want to get a capacity of 40 μF are (alternate plates are connected together):

The correct option is C 9Connecting n plates alternatively creates a capacitance of (n−1)Aϵ0d.Given, capacity of each condenser: ϵ0Ad=5 μFSo,(n−1)Aϵ0d=40 μF⇒n=9

The capacity between two adjacent plates of parallel plate condenser is $5 μ F$ . The number of plates used if we want to get a capacity of $40 μ F$ are (alternate plates are connected together):

The correct option is C $9$
Connecting $n$ plates alternatively creates a capacitance of $\frac{(n - 1) A ϵ_{0}}{d}$ .
Given, capacity of each condenser: $\frac{ϵ_{0} A}{d} = 5 μ F$
So,
$\frac{(n - 1) A ϵ_{0}}{d} = 40 μ F$
$\Rightarrow n = 9$