You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VIII

Physics

Capacitors

The value of ...

Question

The value of capacitance of an air capacitor is $8 μ F$ . Two dielectrics of identical size fill the space between the plates as shown. Dielectric constants are $K_{1} = 3.0$ and $K_{2} = 6.0$ . Calculate the value of the new capacitance.

Open in App

Solution

$C = E_{0} \frac{a}{d} = 8 p f$

$C^{'} = \frac{E_{0} a}{\frac{t_{1}}{r_{1}} + \frac{t_{2}}{r_{2}}} = \frac{E_{0} a}{\frac{\frac{d}{2}}{R_{1}} + \frac{\frac{d}{2}}{R_{2}}} = 2 C [\frac{1}{3} + \frac{1}{6}] = \frac{1}{\frac{1}{2}} 2 C = 4 C$

$4 \times 8 p f = 32 p f$

Suggest Corrections

Similar questions

Q. The space between the plates of a parallel plate capacitor of capacitance

C

is filled with three dielectric slabs of identical size as shown in figure. If the dielectric constants of the three slabs are

K_{1}

K_{2}

and

K_{3}

, find the new capacitance.

Q. Air filled capacitor of capacitance

2 μ F

is filled with three dielectric material of dielectric constants

K_{1} = 4, K_{2} = 4 a n d K_{3} = 6

as shown in the figure. The new capacitance of the capacitors is

Q. Three slabs of same dimension having dielectric constant

K_{1}, K_{2}

and

K_{3}

completely fills the space between the plates of a capacitor as shown in figure. If

C

is the initial capacitance of the capacitor, the new capacitance is?

Q. Two dielectric slabs of constant

K_{1}

and

K_{2}

have been filled in between the plates of a capacitance as shown. What will be the capacitance of the capacitor?

A parallel plate capacitor with air between the plates has a capacitance of $9 p F$ . The separation between its plates is $d$ . The space between the plate is now partially filled with two dielectrics. One of the dielectric has dielectric constant $K_{1} = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $K_{2} = 6$ and thickness $\frac{d}{3}$ . The remaining $\frac{d}{3}$ space is filled with air. The new capacitance is :

Join BYJU'S Learning Program

The value of capacitance of an air capacitor is 8μF. Two dielectrics of identical size fill the space between the plates as shown. Dielectric constants are K1=3.0 and K2=6.0. Calculate the value of the new capacitance.

C=E0ad=8pfC′=E0at1r1+t2r2=E0ad2R1+d2R2=2C[13+16]=1122C=4C4×8pf=32pf

The value of capacitance of an air capacitor is $8 μ F$ . Two dielectrics of identical size fill the space between the plates as shown. Dielectric constants are $K_{1} = 3.0$ and $K_{2} = 6.0$ . Calculate the value of the new capacitance.

$C = E_{0} \frac{a}{d} = 8 p f$

$C^{'} = \frac{E_{0} a}{\frac{t_{1}}{r_{1}} + \frac{t_{2}}{r_{2}}} = \frac{E_{0} a}{\frac{\frac{d}{2}}{R_{1}} + \frac{\frac{d}{2}}{R_{2}}} = 2 C [\frac{1}{3} + \frac{1}{6}] = \frac{1}{\frac{1}{2}} 2 C = 4 C$

$4 \times 8 p f = 32 p f$