Capacitors
14.0 Some important concepts
14.0 Some important concepts
1. Problems based on change in capacity: This can be done by following methods,
- By inserting a dielectric slab or removing the slab (if it already exists) or
- By changing the distance between the plates or area of plates of capacitor.
- By both of above.
Now there are two types of problems these are when
(a) Battery is removed after charging: If battery is removed after charging, then the charge on the plates of capacitor that was stored already, remains constant. Now charge on capacitor is multiplication of voltage and capacity, if capacity of capacitor varies then voltage will also change and it will be inversely proportional that of the capacity.
If initial capacity is $C_i$ and final is $C_f$ then, capacity after inserting the slab of dielectric constant $K$ is, $${C_f} = {C_i}K$$Change in electric field $E$ is, $${E_f} = \frac{{{E_i}}}{K}$$ Change in potential difference $V$ is,$${V_f} = \frac{{{V_i}}}{K}$$ Change in stored potential energy$U$ is, $$\begin{equation} \begin{aligned} U = \frac{1}{2}\frac{{{q^2}}}{C} \\ U\alpha \frac{1}{C} \\ {U_f} = \frac{{{U_i}}}{K} \\\end{aligned} \end{equation} $$
(b) Battery remains connected: When battery remains connected, potential difference across the terminals of capacitor will remain unchanged so the potential difference across the terminals of capacitor will remain constant. So if capacity of capacitor changes the charge will also change accordingly and change in charge will be proportional to that of capacity.
So, If initial and final capacities are $C_i$ and $C_f$ respectively then, $${C_f} = {C_f}K$$ Since electric field is proportional to potential difference $V$,so electric field will remains unchanged. Change in potential energy, since $$\begin{equation} \begin{aligned} U = \frac{1}{2}C{V^2} \\ U\alpha C \\ {U_f} = K{U_i} \\\end{aligned} \end{equation} $$
2. Heat Generation: When the switch in a capacitive circuit is closed or opened, in the other word new connection are made the charge flows in the circuit, it cause heat loss that is called heat generation. The problems of heat generation can be solved by simply energy conservation principle. For this remember that when a charge $+q$ flows from negative terminal to the positive terminal inside a battery of emf $V$ is supplies a energy,$$E = qV$$and when the opposite case occurs, i.e., $+q$ charge flows in opposite direction, then it consumes energy by the same amount. From energy conservation principle we can write that
Heat generated = energy supplied by the battery(/batteries) $-$ energy consumed by the battery (/batteries) $ + \sum {{U_i} - \sum {{U_f}} } $
${\sum {{U_i}} }$= energy stored in the capacitor initially
${\sum {{U_f}} }$= energy stored in the capacitor finally.
3. Dielectric Breakdown: When a dielectric material is subjected to a strong electric field, dielectric becomes a conductor. This occurs when the electric field is so strong that electrons are ripped loose from their molecules. This maximum electric field magnitude that a material can withstand without the occurrence of breakdown is called its dielectric strength.
4. Force on Dielectric slab at constant Potential $V$: When a dielectric slab is being inserted between the plates of capacitor it is attracted into the capacitor. This happen because of the induced charge on the slab of dielectric material. 
There is already some charge on the capacitor and that is just in such a combination that that charge attract the charge on the slab (opposite in nature) as shown in the figure.
There is already some charge on the capacitor and that is just in such a combination that that charge attract the charge on the slab (opposite in nature) as shown in the figure. Now to calculate the force exerted on the slab, we will consider a circuit diagram as shown in the figure including partially inserted slab in a capacitor. Let the width of plates is $b$, length is $l$, and $d$ is the distance between the plates. We want to find force on a slab of dielectric constant $K$ and thickness $d$, when it's $x$ length is inside the plates.
Now what we are going to do is first we will find the equivalent capacitance of the arrangement and then we will find the total energy stored in the capacitor and by differentiating that energy expression with respect to displacement, we will find the negative of the force exerted on the slab. The direction of the force is always inwards whatever slab is inserting or exiting that means force is attracting is nature.
$$C = \frac{{A{\varepsilon _ \circ }}}{d}$$Where $A$= Area = width$ \times $ length $$\begin{equation} \begin{aligned} {C_1} = \frac{{{\varepsilon _ \circ }b(1 - x)}}{d} \\ {C_2} = \frac{{K{\varepsilon _ \circ }bx}}{d} \\\end{aligned} \end{equation} $$Hence the net capacitance of the capacitor is$$C = {C_1} + {C_2}$$$$C = \frac{{{\varepsilon _ \circ }b}}{d}(l + x(K - 1))$$The energy stored in the capacitor at this instant is $$\begin{equation} \begin{aligned} U = \frac{1}{2}C{V^2} \\ dU = \frac{1}{2}({V^2}).dC \\ dC = \frac{{{\varepsilon _ \circ }b}}{d}(K - 1)dx \\ dU = \frac{{{\varepsilon _ \circ }b}}{{2d}}(K - 1){V^2}dx \\\end{aligned} \end{equation} $$Since force is generated due to charges, it is electrostatic in nature and conservative also.So,$$\begin{equation} \begin{aligned} F = - \frac{{dU}}{{dx}} \\ \frac{{dU}}{{dx}} = \frac{{{\varepsilon _ \circ }b}}{{2d}}(K - 1){V^2} \\ F = - \frac{{{\varepsilon _ \circ }b}}{{2d}}(K - 1){V^2} \\\end{aligned} \end{equation} $$Thus, the electric field attracts the dielectric into the capacitor with the force $F$ shown above. Note that the force is constant.It is independent of the distance $x$.
Now we can further extend the result as $$\begin{equation} \begin{aligned} F = - \frac{{{\varepsilon _ \circ }b}}{{2d}}(K - 1){V^2} = ma \\ a = - \frac{{{\varepsilon _ \circ }b}}{{2md}}(K - 1){V^2} \\\end{aligned} \end{equation} $$The equilibrium position of the slab is at the instant when the slab is fully inside the plates. So, the slab will execute oscillations. $T/4$ is the time taken to reach from one end to other end of the capacitor,say it is $t$, Using $$\begin{equation} \begin{aligned} s = \frac{1}{2}a{t^2} \\ t = \sqrt {\frac{{2s}}{a}} \\ t = \sqrt {\frac{{2(l - x)}}{{\frac{{{\varepsilon _ \circ }(A/l){V^2}(K - 1)}}{{2md}}}}} = \sqrt {\frac{{4(l - x)mdl}}{{{\varepsilon _ \circ }(A/l){V^2}(K - 1)}}} \\\end{aligned} \end{equation} $$So the time period,$$T = 4t = 8\sqrt {\frac{{4(l - x)mdl}}{{{\varepsilon _ \circ }(A/l){V^2}(K - 1)}}} $$
