Separation between plates of a parallel plate capacitor is d and area of each plate is A . When a sl

English

Gujarati

Hindi

2. Electric Potential and Capacitance

hard

Separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t(t < d)$ is introduced between the plates, its capacitance becomes

$\frac{{{\varepsilon _0}A}}{{d + t\left( {1 - \frac{1}{k}} \right)}}$

$\frac{{{\varepsilon _0}A}}{{d + t\left( {1 + \frac{1}{k}} \right)}}$

$\frac{{{\varepsilon _0}A}}{{d - t\left( {1 - \frac{1}{k}} \right)}}$

$\frac{{{\varepsilon _0}A}}{{d - t\left( {1 + \frac{1}{k}} \right)}}$

Solution

(c) Potential difference between the plates $V$ = $V_{air}$ + $V_{medium}$
$ = \frac{\sigma }{{{\varepsilon _0}}} \times (d – t) + \frac{\sigma }{{K{\varepsilon _0}}} \times t$
$==>$ $V = \frac{\sigma }{{{\varepsilon _0}}}(d – t + \frac{t}{K})$
$ = \frac{Q}{{A{\varepsilon _0}}}(d – t + \frac{t}{K})$
Hence capacitance $C = \frac{Q}{V} = \frac{Q}{{\frac{Q}{{A{\varepsilon _0}}}(d – t + \frac{t}{K})}}$
$ = \frac{{{\varepsilon _0}A}}{{(d – t + \frac{t}{k})}} = \frac{{{\varepsilon _0}A}}{{d – t\,\left( {1 – \frac{1}{k}} \right)}}$

Standard 12

Physics

When air in a capacitor is replaced by a medium of dielectric constant $K$, the capacity

easy

Two identical parallel plate capacitors are connected in series to a battery of $100\,V$. A dielectric slab of dielectric constant $4.0$ is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively

medium

In a parallel plate capacitor set up, the plate area of capacitor is $2 \,m ^{2}$ and the plates are separated by $1\, m$. If the space between the plates are filled with a dielectric material of thickness $0.5\, m$ and area $2\, m ^{2}$ (see $fig.$) the capacitance of the set-up will be $………\, \varepsilon_{0}$

(Dielectric constant of the material $=3.2$ ) and (Round off to the Nearest Integer)

hard

(JEE MAIN-2021)

In a parallel plate capacitor the separation between the plates is $3\,mm$ with air between them. Now a $1\,mm$ thick layer of a material of dielectric constant $2$ is introduced between the plates due to which the capacity increases. In order to bring its capacity to the original value the separation between the plates must be made……$mm$