Question

A metal plate of thickness half the separation between the capacitor plates of capacitance C is inserted. The new capacitance is:

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

Answer 1

The correct option is D 2C

Initial capacitance $=\frac{{\epsilon }_{0}A}{d}$
Now after metal plate is inserted it behaves as two capacitor plates in series with capacitance $\frac{{\epsilon }_{0}A}{\left(\frac{d}{a}\right)}$
We know when capacitance are in series
$\frac{1}{C_{eff}}=\frac{1}{C_1}+\frac{1}{C_2}$
$\Rightarrow \frac{1}{C_{eff}}=\frac{d}{4{\epsilon }_{0}A}+\frac{d}{4{\epsilon }_{0}A}$
$\Rightarrow C_{eff}=\frac{2{\epsilon }_{0}A}{d.}$
$\therefore Now{C}_{eff}=2C$