The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively C0 and W0 . If the air is replaced by glass (dielectric constant =5 ) between the plates, the capacity of the plates and the energy stored in it will respectively be

Question

The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively C0 and W0 . If the air is replaced by glass (dielectric constant =5 ) between the plates, the capacity of the plates and the energy stored in it will respectively be (A) 5 C0, 5 W0 (B) 5 Co, (W0/5) (C) (Co/5), 5 Wo (D) (Co/5), (Wo/5)

Tardigrade Admin · Accepted Answer

When a dielectric K is introduced in a parallel plate condenser, its capacity becomes K times. Hence C'=5 C0 Energy stored, W0=(q2/2 C0) ∴ W'=(q2/2 C')=(q2/2 × 5 C0) ⇒ W'=(W0/5)

Q. The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively $C_{0}$ and $W_{0} .$ If the air is replaced by glass (dielectric constant $= 5$ ) between the plates, the capacity of the plates and the energy stored in it will respectively be

$5 C_{0}, 5 W_{0}$

$5 C_{o}, \frac{W _{0}}{5}$

$\frac{C _{o}}{5}, 5 W_{o}$

$\frac{C _{o}}{5}, \frac{W _{o}}{5}$

When a dielectric $K$ is introduced in a parallel plate condenser, its capacity becomes $K$ times.
Hence $C^{'} = 5 C_{0}$
Energy stored, $W_{0} = \frac{q ^{2}}{2 C _{0}}$
$∴ W^{'} = \frac{q ^{2}}{2 C ^{'}} = \frac{q ^{2}}{2 \times 5 C _{0}}$
$\Rightarrow W^{'} = \frac{W _{0}}{5}$

Q. The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively C0​ and W0​. If the air is replaced by glass (dielectric constant =5 ) between the plates, the capacity of the plates and the energy stored in it will respectively be

5C0​,5W0​

5Co​,5W0​​

5Co​​,5Wo​

5Co​​,5Wo​​