A parallel plate capacitor with plate area A | vedclass

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

Question

(C) The capacitor is divided into two parts in series,each with a plate separation of $d/2$ and area $A$.Let $C_1$ and $C_2$ be the capacitances of th

Vedclass · Accepted Answer

$\frac{2 \varepsilon_0 A}{d}\left(\frac{K_1 K_2}{K_1+K_2}\right)$

Vedclass · Answer

(C) The capacitor is divided into two parts in series,each with a plate separation of $d/2$ and area $A$.Let $C_1$ and $C_2$ be the capacitances of the two parts:$C_1 = \frac{K_1 \varepsilon_0 A}{d/2} = \frac{2 K_1 \varepsilon_0 A}{d}$$C_2 = \frac{K_2 \varepsilon_0 A}{d/2} = \frac{2 K_2 \varepsilon_0 A}{d}$Since they are in series,the equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{2 K_1 \varepsilon_0 A} + \frac{d}{2 K_2 \varepsilon_0 A}$$\frac{1}{C_{eq}} = \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)$$C_{eq} = \frac{2 \varepsilon_0 A}{d} \left( \frac{K_1 K_2}{K_1 + K_2} \right)$

$A$ parallel plate capacitor with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the figure. The dielectric constants are $K_1$ and $K_2$ respectively. The capacitance will be:

Similar Questions

If the distance between the plates of a parallel plate capacitor is halved and the dielectric constant of the dielectric is doubled,then its capacity will

Two capacitors $C$ and $2C$ are connected in parallel and charged to a potential $V$ by a battery. The battery is then removed. The capacitor $C$ is filled with a dielectric of constant $K$. What is the new potential difference across the capacitors?

The force between two point charges placed in a material medium of dielectric constant $K$ is $F$. If the material is removed,then the force between them becomes . . . . . . .

$A$ parallel plate capacitor,partially filled with a dielectric slab of dielectric constant $K$,is connected to a cell of emf $V \text{ volt}$,as shown in the figure. The separation between the plates is $D$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module