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  4. In a parallel plate capacitor of capacitance C, a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plates. The capacitance now becomes

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Q. In a parallel plate capacitor of capacitance $C$, a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plates. The capacitance now becomes

KCETKCET 2001Electrostatic Potential and Capacitance

Solution:

Method 1 :
Before the metal sheet is inserted, $C=\varepsilon_0 A / d$.
After the sheet is inserted, the system is equivalent to two capacitors in series, each of capacitance $C^{\prime} \varepsilon_0 A(d / 4)=4 C$.
The equivalent capacity is now $2 C$.
Method 2 :
If a dielectric slab of dielectric constant $K$ and thickness $t$ is inserted
in the air gap of a capacitor of plate separation $d$ and plate area $A$, its capacitance becomes
$ C=\varepsilon_0 A / d-t(1-1 / k) . $
Here, the initial capacitance, $C=\varepsilon_0 A / d$
For the metal sheet, $t=d / 2, K=\infty$.
The new capacitance is $ \begin{aligned} C^{\prime} & =\frac{\varepsilon_0 A}{d-\frac{d}{2}\left(1-\frac{1}{\infty}\right)} \\ & =\frac{\varepsilon_0 A}{\frac{d}{2}}=2 \frac{\varepsilon_0 A}{d} \\ & =2 C . \end{aligned} $