Question

A parallel plate capacitor has a capacitance $C$ with cross-sectional area of the plates $A$ and separation between the plates as $d.$ If a copper plate of same area , but of thickness $\frac{d}{2}$ is placed between the plates, then the new capacitance will be-

• A. Halved
• B. Doubled
• C. One fourth
• D. Unchanged

Answer 1

The correct option is B Doubled


When a conducting material of thickness $\frac{d}{2}$ is inserted in a parallel plate capacitor, its effective capacitance becomes,

$\frac{1}{C_{eff}}=\frac{\frac{d}{2}}{K{\epsilon }_{0}A}+\frac{d-\frac{d}{2}}{{\epsilon }_{0}A}$

$\frac{1}{C_{eff}}=\frac{\frac{d}{2}}{{\epsilon }_{0}A}[\because K=\infty ]$

$\therefore C_{eff}=\frac{2{\epsilon }_{0}A}{d}=2C$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.