Capacitors Concept Page - 6

Definition

Dielectric constant

The ratio of permittivity of a medium to the permittivity of free space is known as the dielectric constant or relative permittivity of the medium. It is denoted by εr${\epsilon }_r$ or K$K$.
K=εεo$K={\displaystyle \frac{\epsilon }{{\epsilon }_o}}$
Note:
K≥1$K\ge 1$ where K=1$K=1$ for free space

Formula

Parallel plate capacitor with a dielectric

The capacitance C$C$ of a parallel plate capacitor with single dielectric slab between plates is :
C=Kϵ0Ad=KC0$C={\displaystyle \frac{{Kϵ}_{0}A}{d}}=K{C}_0$ 
where C0$C_0$ is the capacitance without the dielectric.

Formula

Dielectrics in series

1C=1C1+1C2=2k1k2k1+k2C0${\displaystyle \frac{1}{C}}={\displaystyle \frac{1}{C_1}}+{\displaystyle \frac{1}{C_2}}={\displaystyle \frac{2k_1{k}_2}{k_1+k_2}}{C}_0$
where k1$k_1$, k2$k_2$ are the dielectric constants 

Example

Capacitance of parallel plate capacitor with a combination of dielectrics

Example:
In the figure shown, find the effective capacitance across P and Q. (Area of each plate is α$\alpha$)Solution:
The above arrangement acts as the K1$K_1$ capacitor in parallel with K2$K_2$ and K3$K_3$ which are in series.
C1=k1ε0α2d,C2=k2ε0α22d,C3=k3ε0α22d$C_1={\displaystyle \frac{k_1{\epsilon }_0\alpha }{2d}},C_2={\displaystyle \frac{k_2{\epsilon }_0\alpha 2}{2d}},C_3={\displaystyle \frac{k_3{\epsilon }_0\alpha 2}{2d}}$
C2$C_2$ and C3$C_3$ are in series.
1Ceff=1C2+1C3${\displaystyle \frac{1}{C_{eff}}}={\displaystyle \frac{1}{C_2}}+{\displaystyle \frac{1}{C_3}}$
1Ceff=dk2ε0α+dk3ε0α=Ceff=ε0αd(k2k3k2+k3)${\displaystyle \frac{1}{C_{eff}}}={\displaystyle \frac{d}{k_2{\epsilon }_0\alpha }}+{\displaystyle \frac{d}{k_3{\epsilon }_0\alpha }}=C_{eff}={\displaystyle \frac{{\epsilon }_0\alpha }{d}}({\displaystyle \frac{k_2{k}_3}{k_2+k_3}})$
C1$C_1$ is in parallel with Ceff$C_{eff}$
∴$\therefore$ Total capacitance C=k1ε0α2d+εoαd(k2k3k2+k3)$C={\displaystyle \frac{k_1{\epsilon }_0\alpha }{2d}}+{\displaystyle \frac{{\epsilon }_o\alpha }{d}}({\displaystyle \frac{k_2{k}_3}{k_2+k_3}})$
C=ε0αd(k12+k2k3k2+k3)$C={\displaystyle \frac{{\epsilon }_0\alpha }{d}}({\displaystyle \frac{k_1}{2}}+{\displaystyle \frac{k_2{k}_3}{k_2+k_3}})$

Definition

variable permittivity

Example : A parallel plate capacitor having square plates of edge and plate separation d$d$. The gap between the plates is filled with a dielectric of dielectric constant K$K$ which varies parallel to an edge as K=K0+αx$K=K_0+\alpha x$ where K$K$ and α$\alpha$ are constants and x$x$ is the distance from the left end.
Solution: Consider a small strip of width dx$dx$ at a separation x$x$ from the left end with area of adx$adx$. Its capacitance is dC=(K0+αx)ϵ0adxd$dC={\displaystyle \frac{(K_0+\alpha x){ϵ}_{0}adx}{d}}$.
The given capacitor may be divided into such strips with x$x$ varying from 0$0$ to a$a$. All these strips are connected to parallel. The capacitance of the given capacitor is,
C=∫a0(K0+αx)ϵ0adxd=ϵ0a2d(K0+aα2)$C={\int }_0^a{\displaystyle \frac{(K_0+\alpha x){ϵ}_{0}adx}{d}}={\displaystyle \frac{{ϵ}_0{a}^2}{d}}(K_0+{\displaystyle \frac{a\alpha }{2}})$

Formula

Cylindrical capacitor with dielectric

Cl=2πϵ0log(ro/ri)(α1ε1+α2ε2)${\displaystyle \frac{C}{l}}={\displaystyle \frac{2\pi {ϵ}_0}{log(r_o/r_i)}}({\alpha }_1{\epsilon }_1{+\alpha }_2{\epsilon }_2)$
where ro$r_o$, ri$r_i$ are the outer and inner radius of the cylinder respectively, α1${\alpha }_1$, α2${\alpha }_2$ are the filling fractions of ε1${\epsilon }_1$ and ε2${\epsilon }_2$ respectively.

Example

Force on a dielectric slab in a conductor

Example:A parallel plate capacitor is made of two plates of length l$l$, width w$w$ and separated by distance d$d$. A dielectric slab (dielectric constant K$K$) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force F=−∂U∂x${\displaystyle F=-{\displaystyle \frac{\mathrm{\partial }U}{\mathrm{\partial }x}}}$ where U is the energy of the capacitor when dielectric is inside the capacitor up to distance x$x$ (See figure). If the charge on the capacitor is Q then the force on the dielectric when it is near the edge is :

Solution:
Let at any moment of time when slab traveled distance x$x$,
Capacitance due to dielectric slab, 
Cd=KϵoA/d=Kϵo(wx)/d$C_d=K{ϵ}_{o}A/d=K{ϵ}_o(wx)/d$
Capacitance where slab is not there:
C=ϵo(w(l−x))/d${\displaystyle C={ϵ}_o(w(l-x))/d}$
Net capacitance will be:
C′=wϵod(l+(k−1)x)${\displaystyle C^{\prime }={\displaystyle \frac{w{ϵ}_o}{d}}(l+(k-1)x)}$
Now, energy U is:
U=Q2C′=Q2d2wϵol(11+(K−1)xl)${\displaystyle U={\displaystyle \frac{Q^2}{C^{\prime }}}={\displaystyle \frac{Q^{2}d}{2w{ϵ}_{o}l}}({\displaystyle \frac{1}{1+{\displaystyle \frac{(K-1)x}{l}}}})}$
Force, F=−dU/dx$F=-dU/dx$, therefore,
F=Q2d2wϵol2(K−1)${\displaystyle F=\frac{Q^{2}d}{2w{ϵ}_o{l}^2}(K-1)}$

Definition

Electric Displacement

Electric displacement is given as:
D⃗ =εoE⃗ +P⃗ $\overrightarrow{D}={\epsilon }_o\overrightarrow{E}+\overrightarrow{P}$
where:
E⃗ :$\overrightarrow{E}:$ External electric field
P⃗ :$\overrightarrow{P}:$ Polarisation 
εo:${\epsilon }_o:$ Permittivity of free space

It is related to the free charge density by, D⃗ .n^=σ$\overrightarrow{D}.\hat{n}=\sigma$

Note:
1. Electric Displacement is directly related to the free surface charge density in a medium instead of the electric field.
2. On change of medium of an electric field, electric field changes but electric displacement remains the same.

Definition

Relationship between Electric Displacement and Electric field

Electric displacement is given by:
D⃗ =εoKE⃗ $\overrightarrow{D}={\epsilon }_{o}K\overrightarrow{E}$
where
εoK:${\epsilon }_{o}K:$ Permittivity of the medium

Formula

Force between the plates of a capacitor

The capacitance C is given by :
C=ϵoAd$C={\displaystyle \frac{{ϵ}_{o}A}{d}}$
Energy stored in caapcitor is U=12CV2$U={\displaystyle \frac{1}{2}}C{V}^2$
                                                   =AϵoN22d$={\displaystyle \frac{A{ϵ}_o{N}^2}{2d}}$
Differentiating this with respect to distance d:
dUdd=−ϵoAV22d2${\displaystyle \frac{dU}{dd}}=-{\displaystyle \frac{{ϵ}_{o}A{V}^2}{2d^2}}$
Force of attraction=−dUdd$=-{\displaystyle \frac{dU}{dd}}$
and Q=CV, so
F=−Q22ϵ0A

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