The plates of a parallel plate capacitor are | vedclass

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

Question

(C) The capacitance of a parallel plate capacitor with air is $C = \frac{\varepsilon_0 A}{d}$.When a dielectric slab of thickness $t$ and dielectric c

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The capacitance of a parallel plate capacitor with air is $C = \frac{\varepsilon_0 A}{d}$.When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted,the new capacitance is $C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$.Since the potential difference $V$ is maintained constant and the charge $Q$ remains constant (as the battery is disconnected),the capacitance must remain the same: $C = C'$.Therefore,$\frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 A}{d' - t + \frac{t}{K}}$,where $d'$ is the new distance between the plates.This simplifies to $d = d' - t + \frac{t}{K}$,or $d' - d = t(1 - \frac{1}{K})$.Given $d' - d = 3.2 \ mm$ and $t = 4 \ mm$,we have:$3.2 = 4(1 - \frac{1}{K})$$\frac{3.2}{4} = 1 - \frac{1}{K}$$0.8 = 1 - \frac{1}{K}$$\frac{1}{K} = 1 - 0.8 = 0.2$$K = \frac{1}{0.2} = 5$.

Similar Questions

Two metallic plates of radius $r$ are placed at a distance $d$ apart,and the capacitance is $C$. If a plate of radius $r/2$ and thickness $d$ with a dielectric constant $6$ is placed between the plates of the capacitor,then its new capacitance will be

The capacitance of a parallel plate capacitor with air as the medium is $3 \mu F$. With the introduction of a dielectric medium between the plates,the capacitance becomes $15 \mu F$. The permittivity of the medium in $SI$ units is: [Given: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]

The variation of the electric field between the plates of a capacitor is shown in the figure. Find the relation between $K_1$,$K_2$,and $K_3$,if $2x = y$.

$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module