Question

The two plates of a parallel plate capacitor are $4 \,mm$ apart. A slab of dielectric constant $3$ and thickness $3 \,mm$ 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 $d=4 \,mm =0.004 \,m$,
$K =3$, thickness $t =3\, mm =0.003 \,m$ and $d_{1}=$ ?
$\therefore C=\frac{\varepsilon_0 A}{d}$ and $C_1=\frac{\varepsilon_0 A}{d_1-t\left(1-\frac{1}{K}\right)}$
since $C_{1}=\frac{2}{3} C$ (given)
$\therefore \frac{\varepsilon_0 A}{d_1-t\left(1-\frac{1}{K}\right)}=\frac{2}{3} \frac{\varepsilon_0 A}{d}$
$\frac{1}{d_{1}-t\left(1-\frac{1}{K}\right)}=\frac{2}{3 d}$
$\frac{1}{d_{1}-0.003\left(1-\frac{1}{3}\right)}=\frac{2}{3 \times 0.004}$
$\frac{1}{d_{1}-0.003 \times \frac{2}{3}}=\frac{1}{0.006}$
$\frac{1}{d_{1}-0.002}=\frac{1}{0.006}$
$d_{1}-0.002=0.006$
$d_{1}=0.006+0.002=0.008\, m =8\, mm .$