Two dielectric slabs of dielectric constants | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The capacitor can be viewed as two capacitors in series,each with plate area $A$ and plate separation $d/2$.The capacitance of the first capacitor

Vedclass · Accepted Answer

$\frac{{2{\varepsilon _0}A}}{d}\left( {\frac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}} \right)$

Vedclass · Answer

(D) The capacitor can be viewed as two capacitors in series,each with plate area $A$ and plate separation $d/2$.The capacitance of the first capacitor is $C_1 = \frac{K_1 \varepsilon_0 A}{d/2} = \frac{2 K_1 \varepsilon_0 A}{d}$.The capacitance of the second capacitor is $C_2 = \frac{K_2 \varepsilon_0 A}{d/2} = \frac{2 K_2 \varepsilon_0 A}{d}$.Since they are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.$\frac{1}{C_{eq}} = \frac{d}{2 K_1 \varepsilon_0 A} + \frac{d}{2 K_2 \varepsilon_0 A} = \frac{d}{2 \varepsilon_0 A} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) = \frac{d}{2 \varepsilon_0 A} \left( \frac{K_1 + K_2}{K_1 K_2} \right)$.Therefore,$C_{eq} = \frac{2 \varepsilon_0 A}{d} \left( \frac{K_1 K_2}{K_1 + K_2} \right)$.

Two dielectric slabs of dielectric constants $K_1$ and $K_2$ are filled between the plates of a parallel plate capacitor as shown in the figure. If the area of each plate is $A$ and the separation between the plates is $d$,what is the equivalent capacitance of the capacitor?

Similar Questions

$A$ parallel plate capacitor has a plate separation of $0.01\, mm$ and uses a dielectric (whose dielectric strength is $19\, kV/mm$) as an insulator. The maximum potential difference that can be applied to the terminals of the capacitor is......$V$

$A$ capacitor of capacitance $15 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.5$,dielectric strength $30 \, MV/m$,and potential difference $V = 30 \, V$. The area of the plate is ....... $\times 10^{-4} \, m^{2}$.

In a parallel plate capacitor of capacitance $C$,a metal sheet is inserted between the plates,parallel to them. If the thickness of the sheet is half of the separation between the plates,the new capacitance will be:

If a dielectric substance is introduced between the plates of a charged air-gap capacitor,the energy of the capacitor will:

Vedclass Test Series

Exam Paper Generator

Online Exam Module