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}= KC_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}}