Physics > Capacitors > 3.0 Parallel Plate Capacitor
Capacitors
1.0 Introduction
2.0 Different types of capacitors and its capacitance.
3.0 Parallel Plate Capacitor
3.1 When equal and opposite charges placed on plates
3.2 When unequal charges are placed on the two plates
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
9.1 Effect of Dielectric
9.2 Capacitance of a Capacitor Partially Filled with Dielectric
9.3 Quantities after inserting dielectric in a capacitor (fully)
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
3.2 When unequal charges are placed on the two plates
3.2 When unequal charges are placed on the two plates
9.2 Capacitance of a Capacitor Partially Filled with Dielectric
9.3 Quantities after inserting dielectric in a capacitor (fully)
Suppose charges ${q_1}$ and ${q_2}$ are given to the two plates. The resulting charge distribution on the various surfaces is shown in figure. The distribution can be done by following these steps:
Step: $1$ First add the whole charge on the various plates of the system (capacitors) and divide by $2$ and put this amount of charge on the both extreme surfaces ($A$ and $D$) as shown in figure shown below:
Step: $2$ Now let's the charge on the inner surface ($B$) of the first extreme plate is $Q$ and the total charge on the first plate should be ${q_1}$ so, $$\frac{{{q_q} + {q_2}}}{2} + Q = {q_1}$$ $$\begin{equation} \begin{aligned} Q = {q_1} - \frac{{({q_1} + {q_2})}}{2} \\ \\\end{aligned} \end{equation} $$ $$Q = \frac{{{q_1} - {q_2}}}{2}$$
Step: $3$ Now the charge on the surface $C$ should be opposite of surface $B$ i. e.,$$ - Q = - \left( {\frac{{{q_1} - {q_2}}}{2}} \right)$$
that's why because the surface $B$ and $D$ is forming a capacitor.
Step: $4$ Do the same for the all surfaces of all plates and final distribution of the charges are as shown below in the figure.
Note that this method is also applicable for a system having more than one plates, Now the electric field between the plates is,$$E = \frac{\sigma }{{{\varepsilon _ \circ }}} = \frac{{({q_1} - {q_2})}}{{2A{\varepsilon _ \circ }}}$$
Thus, the potential difference between the plates is,$$V = E.d = \frac{{({q_1} - {q_2})d}}{{2A{\varepsilon _ \circ }}}$$
In order to find $C = q/V$ we must now use the value $({q_1} - {q_2})/2$ for $q$ because this is the charge responsible for the field between two plate. The outer charge do not come in picture. Thus,we get $$C = \frac{{{\varepsilon _ \circ }A}}{d}$$
