3.1 When equal and opposite charges placed on plates

  Capacitors

    1.0 Introduction
    2.0 Different types of capacitors and its capacitance.
    3.0 Parallel Plate Capacitor
    4.0 Capacitance of spherical conductor
    5.0 Capacitance of a earthed sphere by a concentric spherical shell
    6.0 Capacitance of a cylindrical capacitor
    7.0 Mechanical force on the charged conductor
    8.0 Redistribution of Charge
    9.0 Dielectrics (Insulators) and Polarization
    10.0 Combination of capacitors
    11.0 Energy Density ($u$)
    12.0 $R$-$C$ Circuits
    13.0 Method of Finding Equivalent Capacitance
    14.0 Some important concepts
    15.0 Van De Graaff Generator

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3.1 When equal and opposite charges placed on plates

We put a charge $q$ on one plate and a charge $-q$ on the other. This can be done either by connecting one plate with the positive terminal and other with the negative of the battery (as shown in figure) or by connecting one plate to the earth and by giving a charge $+q$ to the other plate only.

This charge will induce a charge $-q$ on the earthed plate. The charges will appear on the facing surfaces. The charge density on each of these surface has a magnitude $$\sigma = \frac{q}{A}$$

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If the plates are large as compared to the separation between them, then the electric field between the plates (at points $B$) is uniform and perpendicular to the plates except for a small region near the edge. The magnitude of this uniform field $E$ may be calculated by using the fact that both positive and negative plates produced the electric field in the same direction (from positive plate towards negative plate) of magnitude $$\frac{\sigma }{{2\varepsilon }}$$ and therefore, the net electric field between the plates will be, $$E = \frac{\sigma }{{2{\varepsilon _o}}} + \frac{\sigma }{{2{\varepsilon _o}}} = \frac{\sigma }{{{\varepsilon _o}}}$$

Outside the plates (at point $A$ & $C$) the field due to positive sheet of charge and negative sheet of charge are in opposite direction. Therefore, net field at these points is zero.

The potential difference between the plates is,$$V=E.d=(\frac{\sigma}{\varepsilon_o})d=\frac{{qd}}{{A\varepsilon_o}}$$

The capacitance of the parallel plate capacitor is,$$C=\frac{ q}{V}=\frac{{A\varepsilon_o}}{d}$$

or $$C=\frac{{A\varepsilon_o}}{d}$$

• Note that instead of two plates if there are $n$ similar plates at equal distance each other and alternate plates are connected together, the capacitance of the arrangement is given by,$$C = \frac{{(n - 1)\varepsilon_o A}}{d}$$

• From the above relation it is clear that the capacitance depends only on geometrical factors ($A$ and $d$).