The capacitance of a parallel plate capacitor with plate area A and separation d, is C. The space between the plates is filled with two wedges of dielectric constants K1 and K2 respectively (figure). Find the capacitance of the resulting capacitor.

Question

The capacitance of a parallel plate capacitor with plate area A and separation d, is C. The space between the plates is filled with two wedges of dielectric constants K1 and K2 respectively (figure). Find the capacitance of the resulting capacitor. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let length and breadth of the capacitor be / and b respectively and

d

be the distance between the plates as shown in figure. Then, consider a strip at a distance

x

of width

d x

.
Now,

QR = x tan θ

and

PQ = d - x tan θ

Where

tan θ = d / l

Capacitance of

PQ

C_{1} = \frac{K _{1} ε _{0} ( b d x )}{d - x t a n θ} = \frac{K _{1} ε _{0} ( b d x )}{d - \frac{x d}{l}}

C_{1} = \frac{K _{1} ε _{0} ( b l d x )}{d ( 1 - x )} = \frac{K _{1} ε _{0} A ( d x )}{d ( 1 - x )}

and

C 2 =

capacitance of

QR

= \frac{K _{2} ε _{0} b ( d x )}{x t a n θ}

C_{2} = \frac{K _{2} ε _{0} A ( d x )}{x d} (tan θ = \frac{d}{l})

Now,

C_{1}

and

C_{2}

are in series. Therefore, their resultant capacity CO will be given by

\frac{1}{C _{0}} = \frac{1}{C _{1}} + \frac{1}{C _{2}}

Then,

\frac{1}{C _{0}} = \frac{1}{C _{1}} + \frac{1}{C _{2}}

= \frac{d ( 1 - x )}{K _{1} ε _{0} A ( d x )} + \frac{x . d}{K _{2} ε _{0} A ( d x )}

\frac{1}{C _{0}} = \frac{d}{ε _{0} A ( d x )} (\frac{1 - x}{K _{1}} + \frac{x}{K _{2}})

= \frac{d { K _{2} ( l - x ) + K _{1} x }}{ε _{0} A K _{1} K _{2} ( d x )}

∴ C_{0} = \frac{ε _{0} A K _{1} K _{2}}{d { K _{2} ( l - x ) + K _{1} x }} d x

C_{0} = \frac{ε _{0} A K _{1} K _{2}}{d { K _{2} l + ( K _{1} - K _{2} ) x }} d x

Now, the net capacitance of the given parallel plate capacitor is obtained by adding such infinitesimal capacitors placed parallel from

x = 0

to

x = 1

i.e.

C_{R} = x = 0 \int x = 1 C_{0} = 0 \int 1 \frac{ε _{0} A K _{1} K _{2}}{d { K _{2} l + ( K _{1} - K _{2} ) x }} d x

Finally we get

C_{R} = \frac{K _{1} K _{2} ε _{0} A}{( K _{2} - K _{1} ) d}

In

\frac{K _{2}}{K _{1}}

= \frac{C K _{1} K _{2}}{K _{2} - K _{1}}

In

\frac{K _{2}}{K _{1}},

where

E = \frac{E d}{l}

Q. The capacitance of a parallel plate capacitor with plate area A and separation d, is C. The space between the plates is filled with two wedges of dielectric constants K1​ and K2​ respectively (figure). Find the capacitance of the resulting capacitor.

Q. The capacitance of a parallel plate capacitor with plate area A and separation $d$ , is $C$ . The space between the plates is filled with two wedges of dielectric constants $K_{1}$ and $K_{2}$ respectively (figure). Find the capacitance of the resulting capacitor.