Question

A parallel plate capacitor having an area A and separated by a distance d is filled by a copper plate of area A and thickness b. The new capacitance is:

• A. $\frac{{\epsilon }_{0}A}{d-\frac{b}{2}}$
• B. $\frac{{\epsilon }_{0}A}{2d}$
• C. $\frac{{\epsilon }_{0}A}{d-b}$
• D. $\frac{{\epsilon }_{0}A}{d+\frac{b}{2}}$

Answer 1

The correct option is C $\frac{{\epsilon }_{0}A}{d-b}$

As capacitance $C_0=\frac{{\epsilon }_{0}A}{d}$

$C=\frac{{\epsilon }_{o}A}{[(d-t)+\frac{t}{k}]}$

where A is the area of capacitor

d is distance between plates of capacitor.

t is the thickness of dielectric inserted between the plates .

k is the dielectric constant.

Given t=b and k=∞, for copper plate(metal)
$\therefore$ after inserting copper plate
$C=\frac{{\epsilon }_{0}A}{d-b}$