Two identical metal plates separated by a distance $d$ forms parallel plate capacitor of capacity $C$ . A metal sheet of thickness $d / 2$ and same dimensions is inserted between the plates, so that the air gap is separated into two equal parts. The new capacity of the system will be:

\frac{ε_{0} A}{2 d}

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\frac{ε_{0} A}{4 d}

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\frac{2 ϵ_{0} A}{d}

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\frac{ϵ_{0} A}{d}

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Solution

The correct option is C $\frac{2 ϵ_{0} A}{d}$
The final capacitance in series combination of
$\frac{ε_{0} A}{\frac{d}{4}}$ and $\frac{ε_{0} A}{\frac{d}{4}}$ is
$\frac{1}{C} = \frac{2 d}{4 ϵ_{0} A}$
Hence , $C = \frac{2 ϵ_{0} A}{d}$

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Two identical metal plates separated by a distance d forms parallel plate capacitor of capacity C. A metal sheet of thickness d/2 and same dimensions is inserted between the plates, so that the air gap is separated into two equal parts. The new capacity of the system will be:

The correct option is C 2ϵ0AdThe final capacitance in series combination ofε0 Ad4 and ε0 Ad4 is 1C=2d4ϵ0AHence , C=2ϵ0Ad

Two identical metal plates separated by a distance $d$ forms parallel plate capacitor of capacity $C$ . A metal sheet of thickness $d / 2$ and same dimensions is inserted between the plates, so that the air gap is separated into two equal parts. The new capacity of the system will be:

The correct option is C $\frac{2 ϵ_{0} A}{d}$
The final capacitance in series combination of
$\frac{ε_{0} A}{\frac{d}{4}}$ and $\frac{ε_{0} A}{\frac{d}{4}}$ is
$\frac{1}{C} = \frac{2 d}{4 ϵ_{0} A}$
Hence , $C = \frac{2 ϵ_{0} A}{d}$