Question

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

• A. $\frac{C}{2}$
• B. $\frac{C}{4}$
• C. 4C
• D. 2C

Answer 1

Answer: (D)

Method 1 :
Before the metal sheet is inserted, $C={\epsilon }_{0}A/d$.
After the sheet is inserted, the system is equivalent to two capacitors in series, each of capacitance $C^{\prime }{\epsilon }_{0}A(d/4)=4C$.
The equivalent capacity is now $2C$.
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
$C={\epsilon }_{0}A/d-t(1-1/k).$
Here, the initial capacitance, $C={\epsilon }_{0}A/d$
For the metal sheet, $t=d/2,K=\infty$.
The new capacitance is $\begin{array}{rl}{C}^{\prime }& =\frac{{\epsilon }_{0}A}{d-\frac{d}{2}\left(1-\frac{1}{\infty }\right)}\\ & =\frac{{\epsilon }_{0}A}{\frac{d}{2}}=2\frac{{\epsilon }_{0}A}{d}\\ & =2C.\end{array}$