Capacitors Concept Page - 2

Example

Capacitance of a parallel plate capacitor

 Example - A parallel plate capacitor consists of two metal plates, each of area 150 cm2$c{m}^2$ separated by a vacuum gap d= 0.60 cm thick. What is the capacitance of this device?
What potential difference must be applied between the plates if the capacitor is to hold a charge of magnitude Q=1.0×10−3μC$Q=1.0\times {10}^{-3}\mu C$  on each plate?
The capacitance of a parallel plate capacitor(for air in between) is
C=QV=ϵ0Ad$C={\displaystyle \frac{Q}{V}}={\displaystyle \frac{{ϵ}_{0}A}{d}}$
C=(8.85×10−12)(150×10−4)0.6×10−2$C={\displaystyle \frac{(8.85\times {10}^{-12})(150\times {10}^{-4})}{0.6\times {10}^{-2}}}$
C=2.21×10−11$C=2.21\times {10}^{-11}$
  =22.1pF$=22.1pF$
The voltage difference  between the plates and the magnitude of the charge Q stored on each plate are related via V=QC$V={\displaystyle \frac{Q}{C}}$So V=1.0×10−92.21×10−11$V={\displaystyle \frac{1.0\times {10}^{-9}}{2.21\times {10}^{-11}}}$
      V=45.2volt$V=45.2volt$

Definition

Electric fields for a parallel plate capacitor

For the outer region,
E=σ2ϵ0−σ2ϵ0=0$E={\displaystyle \frac{\sigma }{{2ϵ}_0}}-{\displaystyle \frac{\sigma }{{2ϵ}_0}}=0$
For the inner region i.e. the region between the plates,
E=σ2ϵ0+σ2ϵ0=σϵ0=Qϵ0A$E={\displaystyle \frac{\sigma }{{2ϵ}_0}}+{\displaystyle \frac{\sigma }{{2ϵ}_0}}={\displaystyle \frac{\sigma }{{ϵ}_0}}={\displaystyle \frac{Q}{{ϵ}_{0}A}}$
where σ$\sigma$ is surface charge density, A$A$ is area of cross section.

Example

Redistribution of charges for capacitors brought to a common potential

The capacity of a condenser A is 10μF$10\mu F$ and it is charged using a battery of 100 V$100V$. The battery is disconnected and the condenser A is connected to a condenser B with common potential as 40 V$40V$. The capacity of B is:
Solution:CA=10μF$C_A=10\mu F$
VA=100V$V_A=100V$
So, charge qA=CAVA$q_A=C_A{V}_A$=10×10−6×100$=10\times {10}^{-6}\times 100$ =10−3$={10}^{-3}$

When the battery is disconnected, total charge must be constant.
So, qA+qB=10−3$q_A+q_B={10}^{-3}$Given common potential ΔV=40 V$\mathrm{\Delta }V=40V$
So, qA=CAΔV$q_A=C_A\mathrm{\Delta }V$=10×10−6×40 V=400×10−6 V$=10\times {10}^{-6}\times 40V=400\times {10}^{-6}V$             
And, qB=CBΔV$q_B=C_B\mathrm{\Delta }V$⇒10−3=400×10−6+CB×40$\Rightarrow {10}^{-3}=400\times {10}^{-6}+C_B\times 40$
or, CB=0.6×10−340$C_B={\displaystyle \frac{0.6\times {10}^{-3}}{40}}$=15×10−6 F=15 μF$=15\times {10}^{-6}F=15\mu F$

Definition

Capacitance of an isolated spherical conductor

The potential of a charged conducting sphere is given by V=Q4πϵ0R$V={\displaystyle \frac{Q}{4\pi {ϵ}_{0}R}}$ where R$R$ is the radius of the sphere.
Then C=Q△V=4πϵ01a−1b$C={\displaystyle \frac{Q}{\mathrm{△}V}}={\displaystyle \frac{4\pi {ϵ}_0}{{\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{b}}}}$
Now for an isolated spherical conductor, taking limits as a→R$a\to R$ and b→∞$b\to \mathrm{\infty }$ , we have C=4πϵ0R$C=4\pi {ϵ}_{0}R$

Example

Example of Spherical capacitance

A spherical capacitor has an inner sphere of radius 9 cm and an outer sphere of radius10 cm. The outer sphere is earthed. Assume there is air in the space between the spheres. What is the capacitance of the capacitor?

Given R1=9cmR2=10cm$R_1=9cm{R}_2=10cm$
By using formula of spherical capacitance
C=4πϵ0R1R2R2−R1$C={\displaystyle \frac{4\pi {ϵ}_0{R}_1{R}_2}{R_2-R_1}}$
Putting down all values we will get
C = 100 pF

Definition

Capacitance of spherical conductors

A spherical capacitor consists of a hollow or a solid spherical conductor surrounded by another concentric hollow spherical conductor. The capacitance of a spherical capacitor is derived as :
By Gauss law charge enclosed by gaussian sphere of radius r be
q=ϵoEA$q={ϵ}_{o}EA$
q=ϵoE(4πr2)$q={ϵ}_{o}E(4\pi r^2)$
∴E=q4πϵor2$\therefore E={\displaystyle \frac{q}{4\pi {ϵ}_o{r}^2}}$
V=−∫Edr$V=-\int Edr$
V=−q4πϵo∫R1R2drr2$V={\displaystyle \frac{-q}{4\pi {ϵ}_o}}{\int }_{R_2}^{R_1}{\displaystyle \frac{dr}{r^2}}$

V=q(R2−R1)4πϵo(R1R2)$V={\displaystyle \frac{q(R_2-R_1)}{4\pi {ϵ}_o(R_1{R}_2)}}$
But C=qV$C={\displaystyle \frac{q}{V}}$
C=4πϵ0R1R2R2−R1$C={\displaystyle \frac{4\pi {ϵ}_0{R}_1{R}_2}{R_2-R_1}}$

Definition

Capacitance of cylindrical capacitors

A cylindrical capacitor consists of a hollow or a solid cylindrical conductor surrounded by another concentric hollow spherical cylinder. The capacitance of a cylindrical capacitor can be derived as:
By Gauss law charge enclosed by gaussian cylinder of radius r and length L is
q=ϵoEA$q={ϵ}_{o}EA$
q=ϵoE(2πrL)$q={ϵ}_{o}E(2\pi rL)$
∴E=q2πϵorL$\therefore E={\displaystyle \frac{q}{2\pi {ϵ}_{o}rL}}$
V=−∫Edr$V=-\int Edr$
V=−q2πϵoL∫R1R2drr$V={\displaystyle \frac{-q}{2\pi {ϵ}_{o}L}}{\int }_{R_2}^{R_1}{\displaystyle \frac{dr}{r}}$
V=q2πϵoLlnr2r1$V={\displaystyle \frac{q}{2\pi {ϵ}_{o}L}}ln{\displaystyle \frac{r_2}{r_1}}$
But C=qV$C={\displaystyle \frac{q}{V}}$
C=2πϵ0Lln(R2/R1)$C={\displaystyle \frac{2\pi {ϵ}_{0}L}{{ln(R}_2/R_1)}}$

Formula

Dielectrics in parallel

C=C1+C2=12(k1+k2)C0$C=C_1+C_2={\displaystyle \frac{1}{2}}{(k}_1+k_2)C_0$
where k1$k_1$,k2$k_2$ are the dielectric constants

BookMarks

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