Question

The plates of a parallel plate capacitor are charged upto $100V$.A $2mm$ thick insulator shunt is inserted between the plates . Then to maintain the same potential difference ,the distance between the same potential difference , the distance between the capacitor plates is increased by $1.6mm$.The dielectric constant of the insulator is

• A. $5$
• B. $8$
• C. $4$
• D. $6$

Answer 1

Answer: (A)

Here,$V=100V$ and $t=2mm$
As $V=\frac{q}{c}$ and $V$ and $q$ remain unchanged ,
therefore C must be constant
ie, $C_i=C_f$
$\frac{{\epsilon }_{0}A}{d}=\frac{{\epsilon }_{0}A}{d+1.6-t+\frac{t}{k}}$
$\therefore 1.6-t+\frac{t}{k}=0$
As $t=2mm$,therefore
$1.6-2+\frac{2}{k}=0$
or $\frac{2}{k}=0.4$
$\therefore k=\frac{2}{0.4}=5$