Consider arrangement shown in figure. total energy stored is U₁ when key is closed. Now key K is m

English

Gujarati

Hindi

2. Electric Potential and Capacitance

hard

Consider the arrangement shown in figure. The total energy stored is $U_1$ when key is closed. Now the key $K$ is made off (opened) and two dielectric slabs of relative permittivity ${ \in _r}$ are introduced between the plates of the two capacitors. The slab tightly fit in between the plates. The total energy stored is now $U_2$. Then the ratio of $U_1/U_2$ is

$\frac{{2{ \in _r}}}{{1\, + \, \in _r^2}}$

${ \in _r}$

$\frac{1}{{{ \in _r}}}$

$\frac{{{ \in _r}}}{{1\, + \, \in _r}}$

Solution

$ \mathrm{u}_{1} =\frac{1}{2} \times 2 \mathrm{Cv}^{2}=\mathrm{Cv}^{2} $

${{\rm{u}}_2} = \frac{1}{2}{\rm{C}}{ \in _{\rm{r}}}{{\rm{v}}^2} + \frac{{{{\rm{C}}^2}{{\rm{v}}^2}}}{{2{_{\rm{r}}}{\rm{C}}}}$

$ = \frac{{{ \in _1}{\rm{C}}{{\rm{v}}^2}}}{2} + \frac{{{\rm{C}}{{\rm{v}}^2}}}{{2 \in {\rm{v}}}}$

${{\rm{u}}_2} = \frac{{{\rm{C}}{{\rm{v}}^2}\left[ { \in _{\rm{r}}^2 + 1} \right]}}{{2{ \in _{\rm{r}}}}} \Rightarrow \frac{{{{\rm{u}}_1}}}{{{{\rm{u}}_2}}} = \frac{{2{ \in _{\rm{r}}}}}{{1 + \in _1^2}}$

Standard 12

Physics

A parallel plate capacitor has plate of length $'l',$ width $'w'$ and separation of plates is $'d'.$ It is connected to a battery of emf $V$. A dielectric slab of the same thickness '$d$' and of dielectric constant $k =4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored$?$

hard

(JEE MAIN-2020)

A parallel plate capacitor with plate area $'A'$ and distance of separation $'d'$ is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :

$\varepsilon(x)=\varepsilon_{0}+k x, \text { for }\left(0\,<\,x \leq \frac{d}{2}\right)$

$\varepsilon(x)=\varepsilon_{0}+k(d-x)$, for $\left(\frac{d}{2} \leq x \leq d\right)$

hard

(JEE MAIN-2021)

The electric field between the plates of a parallel plate capacitor when connected to a certain battery is ${E_0}$. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant $K$ without disturbing the battery connections, the field between the plates shall be

easy

Initially the circuit is in steady state. Now one of the capacitor is filled with dielectric of dielectric constant $2$ . Find the heat loss in the circuit due to insertion of dielectric

hard

A parallel plate capacitor with air as medium between the plates has a capacitance of $10\,\mu F$. The area of capacitor is divided into two equal halves and filled with two media as shown in the figure having dielectric constant ${k_1} = 2$and ${k_2} = 4$. The capacitance of the system will now be…….$\mu F$

medium

Solution

Similar Questions

Initially the circuit is in steady state. Now one of the capacitor is filled with dielectric of dielectric constant $2$ . Find the heat loss in the circuit due to insertion of dielectric

Start a Free Trial Now