Question

The capacitance of parallel plate capacitor is $16\mu F$. When a glass slab is placed between the plates, the potential difference reduces to $1/8^{th}$ of the original value. What is dielectric constant of glass?

• A. $4$
• B. $8$
• C. $16$
• D. $32$

Answer 1

The correct option is B $8$

Given,

$16\mu F$

$V=\frac{Q}{C}=\frac{Q}{16\mu F}$

When dielectric is inserted

$V^{\prime }=\frac{Q}{C}=\frac{Q}{K\left(16\mu F\right)}=\frac{V}{8}$

$\Rightarrow \frac{Q}{K\left(16\mu F\right)}=\frac{Q}{8\times 16\mu F}$

$\Rightarrow K=8$

$\therefore$ Option $B$ is correct.