1. Electric Capacitance:$\\$

• Capacitors are charge storing device.
• Capacitance (C) of a capacitor is the ratio of charge (Q) given and the potential ( V ) to which it is raised i.e $C=\frac{Q}{V}$
• Its S.I unit is Farad and its cgs unit is Statfarad. Stat-farad =$\frac{1}{9}×10^{-11}$ farad
• The dimensional formula of electrical capacitance is $[M^{-1} L^{-2} T^{-4} A^{-2}]$
• The capacitance of a spherical conductor is C=$4πε_0$ R=$\frac{R}{9×10^9}$

where R is the radius of the conductor.

• The capacitance of the earth is $(6.37×10^6 )/(9×10^9 )=7.12×10^{-4}$ farad

## Capacitor and Capacitance

1. Capacitor and Capacitance of Parallel Plate Capacitor\\$

2. A capacitor consists of two conductors separated by an insulator or dielectric. The conductors carry equal and opposite charge±Q. It is an electrical device that stores electric charge.

3. Capacitance of parallel plate capacitor with air as dielectric is$\\$$C_0=\frac{ε_0.A}{d}$ Where A is the area of each plate and 'd’ is the distance between them.

4. With dielectric of dielectric constant $ε_r$ or K. $C_0=\frac{K.ε_0.A}{d}$

5. The capacitance does not depend on the material of plates, potential difference between the plates and the charge given to the plates.

6. The value of C depends on size, shape and relative position of two coatings of the capacitor. It also depends on nature of medium separating the two coatings.

7. A parallel plate capacitor is used to produce uniform electric field between the plates. Outside the plates no electric field is produced. A material used as a dielectric in capacitor must have high dielectric constant and high dielectric strength

8. Capacitance of a parallel plate capacitor with a dielectric slab of dielectric constant K of thickness t(t<d) is given by$\\$C=$\frac{ε_0 .A}{ d-t(1-\frac{1}{K})}$

9. Capacitance of a parallel plate capacitor with a conducting slab of thickness (t<d) is given by C=$\frac{ε_0 .A}{ d-t(1-\frac{1}{∞})}$ [∵K" for " Conductor = ∞] Hence C=$\frac{ε_0.A}{d-t}$

10. If dielectric slab fills the entire space between the plates (instead of filling it partially) i.e. t=d, then C=$\frac{ε_0.A}{\frac{d}{K}}$=K$\frac{ε_0.A}{d-t}$= K$C_0$ If the conducting slab entirely fills the space between plate i.e. t=d then C=$\frac{ε_0.A}{d-t}$=C=$\frac{ε_0.A}{d-d}$ = ∞

11. The capacitance of a parallel plate capacitor having a number of slabs of thickness $t_1,t_2,t_3,……….$ and dielectric constants $K_1,K_2,K_3,………$ respectively in

d=t_{1}+t_{2}+t_{3}+\cdots \ldots \ldots

When two dielectrics of equal thickness are arranged in series then:

\begin{array}{r} C_{\mathrm{eq}}=\frac{\varepsilon_{0} A}{\frac{d / 2}{K_{1}}+\frac{d / 2}{K_{2}}}=\frac{\varepsilon_{o} A}{d}\left(\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}\right) \\ \text { Similarly, } K_{\text {eq }}=\frac{d}{\frac{t_{1}}{K_{1}}+\frac{t_{2}}{K_{2}}+\cdots \cdot \frac{t_{0}}{K_{n}}} \end{array}

For two dielectrics of equal thickness $\left(t_{1}=t_{2}=d / 2\right)$ arranged in series

K_{e q}=\frac{2 K_{1} K_{2}}{K_{I}+K_{2}}