Capacitor problem

Problem:
A capacitor is constructed from two square metal plates of area $L^2$ separated by a distance $d$. One half of the space between the plates is filled with a substance of dielectric constant $(\kappa_1)$. The other half is filled with another substance with constant $(\kappa_2)$. Calculate the capacitance of the device assuming that the free space capacitance is $C_0$.

Solution:
Given
$A = L^2$
$d_1 = d_2 = \frac{d}{2}$
Dielectric constant $\kappa_1$ and $\kappa_1$
$C_0$

Find $C$

Formula
In series $C = \frac{A\epsilon_1\epsilon_2}{\epsilon_{2}d_1+\epsilon_{1}d_2}$
$\epsilon = \epsilon_{0}\kappa$
$C_0 = \epsilon_{0}\frac{A}{d}$

Calculation
$C = \frac{A\epsilon_1\epsilon_2}{\epsilon_{2}d_1+\epsilon_{1}d_2} = \frac{A\epsilon_0\kappa_1\epsilon_0\kappa_2}{(\epsilon_0\kappa_2+\epsilon_0\kappa_1)d/2}$
$C = 2\frac{A\epsilon_0}{d}\frac{\kappa_1\kappa_2}{(\kappa_2+\kappa_1)}$
$C = 2C_0\bigg(\frac{\kappa_1\kappa_2}{\kappa_2+\kappa_1}\bigg)$

Lagrangian problem

Question The Lagrangian for a mechanical system is $L=a\dot q^2+bq^4$ where $q$ is a generalized coordinate and $a$ and $b$ are constants. The equation of motion for this system is
A. $\dot q=\sqrt{\frac{b}{a}}q^2$
B. $\dot q=\frac{2b}{a} q^3$
C. $\ddot q=\frac{-2b}{a} q^3$
D. $\ddot q=+\frac{2b}{a} q^3$
E. $\ddot q=\frac{b}{a} q^3$
Solution:
The Lagrangian equation of motion for the generalized coordinate $q$:
$\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}$
$\frac{\partial L}{\partial q}=\frac{\partial}{\partial q} (a\dot q^2+bq^4 )=0+\frac{\partial}{\partial q} bq^4=4bq^3$
$\frac{\partial L}{\partial \dot q}=\frac{\partial}{\partial \dot q} (a\dot q^2+bq^4 )=\frac{\partial}{\partial \dot q} a\dot q^2+0=2a\dot q$
$\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}=\frac{\partial}{\partial t} 2a\dot q=2a \ddot q$
$\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q} \rightarrow 4bq^3=2a\ddot q \rightarrow \ddot q=\frac{2b}{a} q^3$
source :-http://physics-problems-solutions.blogspot.com

What are constraints

Sometimes motion of a particle or system of particles is restricted by one or more conditions. The limitations on the motion of the system are called constraints. The number of co-ordinates needed to specify the dynamical system becomes smaller when constraints are present in the system. Hence the degree of freedom of a dynamical system is defined as the minimum number of independent co-ordinates required to simplify the system completely along with the constraints. Thus if k is the number of constraints and N is the number of particles in the system possessing motion in three dimensions then the number of degrees of freedom are given by
n=3N-k                                  (1)
thus the above system has n degrees of freedom.
Constraints may be classified in many ways. If the condition of constraints can be expressed as equations connecting the co-ordinates of the particles and possibly the time having the form
f(r₁,r₂,......t)=0                          (2)
then constraints are said to be holonoic and the simplest example of holonomic constraints is rigid body. In case of rigid body motion the distance between any two particles of the body remains fixed and do not change with the tie. If r_i and r_j are the position vectors of the i'th and j'th particles then , the distance between the is given by
|r_i-r_j|=c_ij                                   (3)
The constraints which are not expressible in the form of equation 2 are called non-holonoic for example, the motion of a particle placed on the surface of a sphere of radius a will be described as
|r|≥a or, r-a≥0
Constraints can further be described as (i) rehonoic and(ii) scleronoous. In rehonomous constraints the equation of constraints contains time as explicit variable while in case of scleronomous constraints they are not explicitly dependent on time.

What are Legendre transformations

The transition from Lagrangian to Hamiltonian formulism corresponds to changing the variables in our mechanical functions from $(q,\stackrel{}{\stackrel{.}{q}},t)\stackrel{}{\mathrm{to}(q,p,t)}$

where , p is related to q and $\stackrel{.}{q}$ by the equation

$p_i=\frac{\partial L(q_j,{\dot{q}}_j,t)\stackrel{}{{}_{}}}{\partial {\dot{q}}_i}$

The procedures for switching variables in this manner is provided by the legendre transformations.

Consider a function of only two variables f(x,y), so that differential of f has the form

df=udx+vdy

where , $u=\frac{\mathrm{df}}{\mathrm{dx}}$ and $v=\frac{\mathrm{df}}{\mathrm{dy}}$                      (1)

To change the basis of description from x,y to a new set of variables u,y , so that differential quantities are expressed in terms of differential du and dy. Let g be the function of u and y defined by the equation

g=f-ux

differential of g is given as

dg=df-udx-xdu

or,

dg=vdy-xdu

which is exactly in the desired form. The quantities x and v are now functions of variables u and y given by the relations

$x=-\frac{\partial g}{\partial u},v=\frac{\partial g}{\partial y}$

which are exactly converse of equation 1