A capacitor has air as dielectric medium and two conducting plates of area 12 and they are 0.6 cm apart. When a slab of dielectric having area 12 and 0.6 cm thickness is inserted between the plates, one of the conducting plates has to be moved by 0.2 cm

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A capacitor has air as dielectric medium and two conducting plates of area 12 ${cm}^{2}$ and they are 0.6 cm apart. When a slab of dielectric having area 12 ${cm}^{2}$ and 0.6 cm thickness is inserted between the plates, one of the conducting plates has to be moved by 0.2 cm to keep the capacitance same as in previous case. The dielectric constant of the slab is

(Given $ε_{0} = 8.834 \times 10^{- 12}$ F/m) [2024]

1 1

2 0.66

3 1.33

4 1.50

Ans.
(4)

Parallel plate capacitance, $C = \frac{ϵ_{o} A}{d_{1}}$

For partially filled capacity, $C = \frac{ϵ_{o} A}{(d_{2} - t + \frac{t}{k})}$

For case (i) $C_{1} = \frac{ϵ_{o} A}{0.6}$ , For case (ii) $C_{2} = \frac{ϵ_{o} A}{0.8 - 0.6 + \frac{0.6}{K}}$

Now $C_{1} = C_{2}, \frac{ϵ_{o} A}{0.6} = \frac{ϵ_{o} A}{0.2 + \frac{0.6}{K}}$ ,

$0.6 = 0.2 + \frac{0.6}{K} \Rightarrow K = \frac{3}{2}$

Want Us to Email you About Special Offers & Updates?