A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now an insulating slab of same area but thickness d/2 is inserted between the plates as shown in figure having dielectric constant K(= 4). The ratio of new capacitance to its original capacitance will be,

Question

A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now an insulating slab of same area but thickness d/2 is inserted between the plates as shown in figure having dielectric constant K(= 4). The ratio of new capacitance to its original capacitance will be, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f3eec263bcb455a1b0646de83e9e6076-.png" /> (A) 2: 1 (B) 8: 5 (C) 6: 5 (D) 4: 1

Tardigrade Admin · Accepted Answer

Ca=(∈0 A/d) Ck=(∈0 A/d-t+(t)k) Ck=(∈0 A/d-(d)2+(d)8 C/k)=(8/5)(∈0 A/d) Ck=(8/5)Ca (Ck/Ca)=(8/5)

$2 : 1$

$8 : 5$

$6 : 5$

$4 : 1$

$C_{a} = \frac{\in _{0} A}{d}$
$C_{k} = \frac{\in _{0} A}{d - t + \frac{t}{k}}$
$C_{k} = \frac{\in _{0} A}{d - \frac{d}{2} + \frac{d}{8}}$
$C_{k} = \frac{8}{5} \frac{\in _{0} A}{d}$
$C_{k} = \frac{8}{5} C_{a}$
$\frac{C _{k}}{C _{a}} = \frac{8}{5}$

2:1

8:5

6:5

4:1

Ca​=d∈0​A​ Ck​=d−t+kt​∈0​A​ Ck​=d−2d​+8d​∈0​A​ Ck​=58​d∈0​A​ Ck​=58​Ca​ Ca​Ck​​=58​

$2 : 1$

$8 : 5$

$6 : 5$

$4 : 1$

$C_{a} = \frac{\in _{0} A}{d}$
$C_{k} = \frac{\in _{0} A}{d - t + \frac{t}{k}}$
$C_{k} = \frac{\in _{0} A}{d - \frac{d}{2} + \frac{d}{8}}$
$C_{k} = \frac{8}{5} \frac{\in _{0} A}{d}$
$C_{k} = \frac{8}{5} C_{a}$
$\frac{C _{k}}{C _{a}} = \frac{8}{5}$