Question

The two plates of a parallel plate capacitor are $4mm$ apart. A slab of dielectric constant $3$ and thickness $3mm$ is introduced between the plates with its faces parallel to them. The distance between the plates is so adjusted that the capacitance of the capacitor becomes $2/3^{rd}$ of its original value. What is the new distance between the plates?

• A. 9 mm
• B. 21 mm
• C. 5 mm
• D. 8 mm

Answer 1

Answer: (D)

Here, distance between parallel plates $d=4mm=0.004m$,
$K=3$, thickness $t=3mm=0.003m$ and $d_1=$ ?
$\therefore C=\frac{{\epsilon }_{0}A}{d}$ and $C_1=\frac{{\epsilon }_{0}A}{d_1-t\left(1-\frac{1}{K}\right)}$
since $C_1=\frac{2}{3}C$ (given)
$\therefore \frac{{\epsilon }_{0}A}{d_1-t\left(1-\frac{1}{K}\right)}=\frac{2}{3}\frac{{\epsilon }_{0}A}{d}$
$\frac{1}{d_1-t\left(1-\frac{1}{K}\right)}=\frac{2}{3d}$
$\frac{1}{d_1-0.003\left(1-\frac{1}{3}\right)}=\frac{2}{3\times 0.004}$
$\frac{1}{d_1-0.003\times \frac{2}{3}}=\frac{1}{0.006}$
$\frac{1}{d_1-0.002}=\frac{1}{0.006}$
$d_1-0.002=0.006$
$d_1=0.006+0.002=0.008m=8mm.$