Cartesian to Spherical coordinate system - Coordinate Transformation

In general orthogonal curvilinear coordinate system, general position in 3 dimension is given by, $$\vec{s} = \vec{s}(u_1, u_2, u_3)$$ and small displacement "ds" is can be written as, $$ \vec{ds} = \sum_i^3 \frac{\partial{\vec{s}}}{\partial{u_i}} du_i $$ where $ \frac{\partial{\vec{s}}}{\partial{u_i}} = h_i \hat{e_i} $ and $\hat{e_i} $ is the unit vector along the 'i'th direction. Rewriting it in simple form, we have, $$ \vec{ds} = \sum_i^3 h_i du_i \hat{e_i} $$
For cartesian coordinate system, $h_i = 1$
$\rightarrow$ 
$$ \vec{ds} = dx \hat{e_x} + dy \hat{e_y} + dz \hat{e_z} $$
But there is no unique choice of coordinate system, we can also choose spherical coordinate system as, 
$$ \vec{ds} = dr \hat{e_r} + r \,d\theta\hat{e_\theta} + r \,sin\theta \,d\phi \hat{e_\phi} $$ with corresponding scaling factors. 

To go from one coordinate system to another, we use the relations, $$ x = x(r,\theta,\phi) \\~\\ y = y(r,\theta,\phi) \\~\\ z=z(r,\theta, \phi) $$ 
The only relation we know from our conventional assumption is that, $$ x = \,r \,sin\theta \,cos\phi \\~\\ y = \,r \,sin\theta \,sin\phi \\~\\ z = \,r \,cos\theta\, $$ 
Using the chain rule, $$ dx = \frac{\partial{x}}{\partial{r}}dr + \frac{\partial{x}}{\partial{\theta}} d\theta + \frac{\partial{x}}{\partial{\phi}}d\phi $$
$\rightarrow$ $$ dx = \,sin\theta \,cos\phi \,dr + r \,cos\theta\, cos\phi \,d\theta - r \,sin\theta \,sin\phi \,d\phi $$
Similarly, $$ dy = \, sin\theta \, sin\phi \, dr + r \, cos\theta \, sin\phi \, d\theta + \, r\, sin\theta\, cos\phi \, d\phi $$ and $$ dz = cos\theta \, dr - r\, sin\theta \, d\theta $$
Applying it in our Cartesian equation for "ds", we get, $$ ds =(\,sin\theta \,cos\phi \,dr + r \,cos\theta\, cos\phi \,d\theta - r \,sin\theta \,sin\phi \,d\phi )\hat{e_x} \\~\\+ (\, sin\theta \, sin\phi \, dr + r \, cos\theta \, sin\phi \, d\theta + \, r\, sin\theta\, cos\phi \, d\phi)\hat{e_y} \\~\\+ (cos\theta \, dr - r\, sin\theta \, d\theta) \hat{e_z} $$  

Thus we expressed the infinitesimal displacement in cartesian system using spherical measurements such $ r, \theta, \phi $. Now, from our definition of unit vector, $$ \hat{e_r} = \frac{{\frac{\partial{\vec{s}}}{\partial{r}}}}{|\frac{\partial{\vec{s}}}{\partial{r}}|} $$ and $$ \hat{e_\theta} = \frac{{\frac{\partial{\vec{s}}}{\partial{\theta}}}}{|\frac{\partial{\vec{s}}}{\partial{\theta}}|}$$ and $$\hat{e_\phi} = \frac{{\frac{\partial{\vec{s}}}{\partial{\phi}}}}{|\frac{\partial{\vec{s}}}{\partial{\phi}}|}$$ 
Using the above, equation, we can write the unit vectors along $r,\,\theta,\,\phi $ directions as follows, $$ \hat{e_r} = sin\theta\, cos\phi\, \hat{e_x} \,+ \,\sin\theta\,sin\phi\,\hat{e_y} \,+\, cos\theta\,\hat{e_z} $$ and $$ \hat{e_\theta} = cos\theta\, cos\phi \,\hat{e_x} \,+\, \cos\theta\, sin\phi \,\hat{e_y} - sin\theta\,\hat{e_z} $$ and $$ \hat{e_\phi} = \,-sin\phi \,\hat{e_x} \,+ \,cos\phi\,\hat{e_y} $$ From these three unit vectors we can solve for the other three unit vectors as following,
Solving the first two,
 $$ sin\theta\,\hat{e_r} + \, cos\theta\,\hat{e_\theta} = \,cos\phi \hat{e_x} + \,sin\phi \,\hat{e_y} $$

Combining with the third we can solve for x and y, $$ \hat{e_x} = sin\theta\,cos\phi\,\hat{e_r} + cos\theta\,cos\phi\,\hat{e_\theta} - \,sin\phi\,\hat{e_\phi} $$ and $$ \hat{e_y} = \,sin\theta\,sin\phi \hat{e_r} +\, cos\theta\, sin\phi\, \hat{e_\theta} + \,cos\phi\, \hat{e_\phi} $$ and finally solving for z, $$ \hat{e_z} = cos\theta\,\hat{e_r} - sin\theta\,\hat{e_\theta} $$

That is all we do need to derive for the coordinate transformation from cartesian to spherical. 

Hamilton's equations from variational principle

As we used to derived the Lagrange's equations of motion from variational principle i.e.Hamilton's principle, we can derive Hamilton's equations. 

Hamilton's principle i.e. variational principle gives the condition that, $$ \delta \int_{t_1}^{t_2} L \, dt = 0 $$ 
where L is the Lagrangian of the given system. In the previous derivation, it was considered various paths in the configuration space. Where else, now we will working in the paths which is the part of Phase space where q and p are the independent coordinates.
q - position and 
p - momentum

To go from Lagrangian to Hamilton, we just express the Lagrangian in terms of Hamiltonian where everything is the function of q and p.

$$ \delta \int_{t_1}^{t_2} \left(p_i\dot{q_i} - H (q,p,t)\right) dt = 0 $$  
Let us call this new function as L',
therefore, $$ L' = p_i\dot{q_i} - H $$ and this new L' obeys the Lagrangian equations of motion that is derived earlier from the definition. 
Therefore, $$ \delta \int_{t_1}^{t_2} L'(q, \dot{q}, p, \dot{p}) \,dt = 0 $$
where the integral is defined over 2n dimensional phase space. And it gives symmetrically two sets of relations,
$$ \frac{d}{dt}\left(\frac{\partial{L'}}{\partial\dot{q_j}}\right) - \frac{\partial{L'}}{\partial{q_j}} = 0  \,\,\,\,j = 1,2,...n $$ and
$$ \frac{d}{dt}\left(\frac{\partial{L'}}{\partial\dot{p_j}}\right) - \frac{\partial{L'}}{\partial{p_j}} = 0 \,\,\,\, j = 1,2,....n $$ 
Substituting for L' we get, $$ \frac{\partial{H}}{\partial{q_j}} = -\dot{p_j} $$  and $$ \frac{\partial{H}}{\partial{p_j}} = \dot{q_j} $$ 
which is our desired Hamilton's equations of motion. 
The derivation is nothing but just the result of our Legendre transformation in variational principle. 

Energy stored in the Capacitor and the effect of Dielectric

A capacitor is usually charged by connecting it to a battery. To charge up the capacitor, we take a small elemental positive charge "+dq" from a neutral system and move it a distance "d" so, that the system gets "-dq". The work is done against the Electric field which directs opposite to the motion. 
Let say, the amount of charge piled in the positive plate is +q, so that the potential difference between the plates is "q/C". 
Now, the work required to do move the next "dq" charge is simply just the potential multiplied by the amount of charge [The definition of potential is work done per unit charge].
So, $$ dW = \frac{q}{C} dq $$ 
Then the total work done to charge up the plate to +Q charge is, $$ W = \int_0^Q \frac{q}{C} dq = \frac{Q^2}{2C} $$ Using the relation, $ Q =  CV $ we get, $$ W = \frac{1}{2} CV^2 $$ which is the electrostatic potential energy stored in the system. 

For example, in the case of parallel plate capacitor, the energy stored can be calculated as, where $ C = \frac{\epsilon_0 A}{d} $ and V = Ed ,$$ W = \frac{CV^2}{2} = \frac{\epsilon_0 A E^2 d }{2} $$ 
Here, "Ad" represents the volume of the in between region where the energy is stored. So, we can define a new term as, 
Electrostatic Energy density = $\frac{1}{2} \epsilon_0 E^2 $ 

As an example, if we try to calculate the electrostatic energy density of air at the break down voltage which is about $ E = 3\times 10^6 Vm^{-1} $ and the energy density can be calculated as, 39.825 $Jm^{-3} $, the value is so high. 
One Joule is defined as the work required to move one kilogram object to one meter distance. It is nearly 40 Joules of energy in a volume one meter cube. 

Now, we will give a slight attention to dielectric system. We know there are two types of charges are produced in polarization. First one is the bound charge, $ \rho_b = - \nabla\cdot\vec{P} $ and the surface charge, $ \sigma_b = \vec{P}\cdot \hat{n} $ 
To apply Gauss law, the total charge inside the material is identified as the free charge and the bound charge. $$ \rho _{total}= \rho_{bound} + \rho_{free} $$ 
Using gauss law, $$ \rho_{total} = \epsilon_0\nabla\cdot\vec{E} = -\nabla\cdot\vec{P} +\rho_{free} $$
Calling, $$ \vec{D}= \epsilon_0\vec{E} + \vec{P} $$ we get, $$ \nabla\cdot\vec{D} = \rho_{free} $$ 
In most of the cases, when Electric field is not so much high, Polarization is given by, $$ \vec{P} = \epsilon_0 \chi_e \vec{E} $$
where $ \chi_e $ is called the electric susceptibility. 
 Then, $$ \vec{D} = \epsilon_0 (1+\chi_e) \vec{E}$$ $$\vec{D}= \epsilon \vec{E} $$ and $$ \epsilon_r = \frac{\epsilon}{\epsilon_0}$$ is called relative permitivity or dielectric constant. 

For vacuum, $$ \vec{D} = \epsilon_0 \vec{E_{ext}} $$ since there is no polarization to take place. In a dielectric medium $$ \vec{D} = \epsilon \vec{E_{in}} = \epsilon_r\epsilon_0 \vec{E_{in}} = \epsilon_0 \vec{E_{ext}} $$
Thus it gives, $$ \vec{E_{ext}} = \epsilon_r \vec{E_{in}} \rightarrow \vec{E_{in}} = \frac{\vec{E_{ext}}}{\epsilon_r} $$ which says that the Electric field is reduced to "$\frac{1}{\epsilon_r} $" factors than the Electric field applied (in the vacuum). 

Then, we can see that, if a dielectric medium is introduced in the between region of capacitor plates, capacitance becomes $$ C'= \frac{Q'}{V'} = \frac{Q}{V'}$$ since the charge is the same in both the cases. But, $$ V' = \frac{E'}{d} = \frac{E}{\epsilon_r{d}} $$ and $$ C' = \epsilon_r C_{vacuum} $$
which means the capacitance of the capacitor is increased by the factor of dielectric constant value of the inserted dielectric medium.

Using this, we can find the expression for the new capacitance of a parallel plate capacitor where dielectric material of thickness "t" is introduced. 
To find the new potential difference, $$ V' = - \int_-^+ \vec{E}\cdot \vec{dl} = \int_0^{d-t}E_{vacuum} dl + \int_0^t E_{dielectric} dl $$
which gives, $$ V' = E_{vacuum}(d-t) + \frac{E_{vacuum}}{\epsilon_r} t $$
Electric field between the plates is given by, $$ E_{vacuum} = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0} $$
So, $$ V' = \frac{Q}{A\epsilon_0} [ (d-t) + \frac{t}{\epsilon_r}] = \frac{Q}{A\epsilon_0} [d-t(1-\frac{1}{\epsilon_r})]$$
And the capacitance is given by, $$ C' = \frac{Q}{V'} = \frac{A\epsilon_0}{d-t(1-\frac{1}{\epsilon_r})} $$    

Capacitance for different types of capacitors [Parallel plate, cylindrical and spherical capacitors]

A capacitor is the combination of two metal plates (conductors) having opposite charges separated by distance "d" apart. Don't assume that the metal plates should be rectangular plates with some infinite distance. 
Our usual capacitor is just so small enough within the size of a finger, where the conducting material rolled down in a cylindrical shape with some non conducting or dielectric material in between them. 

Let us discuss with mathematics, 
The two plates have opposite charges, let say, "+Q" on the first plate and "-Q" on the second plate. So, the potential difference between these plates can be calculated as the work done required to bring a unit positive charge from -Q to +Q [from the definition of potential] 
$$ \int_{V_-}^{V_+}\,dV = V = V_+ - V_- = - \int_{-}^{+} \vec{E}\cdot \vec{dl} \,\,\,...eq.(1)$$ 
We don't assume anything about the Shapes of the plates, so Electric field is just given by the definition as, $$ \vec{E} = \frac{1}{4\pi\epsilon_0} \int \rho \frac{\hat{r}}{r^2} d\tau $$ where $\rho$ is the volume charge density and integral is over the volume "V".

Electric field is proportional to both Potential "V" and the charge on the each capacitor "Q" and the ratio is some constant which is defined as the capacitance. $$ C = \frac{Q}{V} $$

Let us try to calculate the capacitance of some simple known shapes, where assumptions are easy to make. 
Capacitance of Parallel Plate Capacitor

First one is the parallel plate capacitor, where Electric field is directed from positive charge to negative charge plate and assumed to be uniform in the direction. So, it can be taken out from the integral in equation (1) and integral sign is canceled by the negative sign and the equation becomes, $$ V = E \int_0^d \,dl = Ed $$ From our definition of Capacitance, $$ C = \frac{Q}{V} = {Q}{E d} $$ And, the Electric field in the region between parallel plate capacitors is given by, $$ E = \frac{\sigma}{\epsilon_0} $$ where $\sigma = \frac{Q}{A}$ is the surface charge density. 
Then, $$ C = \frac{Q}{\frac{Qd}{A\epsilon_0}} = \frac{A\epsilon_0}{d} $$ which is determined only by the sizes, shapes, and separation distance of the two conductors. 

Capacitance of Cylindrical capacitor

Second is the cylindrical capacitors, where inside is solid cylinder with positive charge with radius "a" and outside is hollow cylinder with bigger radius "b" (b>a). Let us take a point "r" in the "in between region a<r<b "
The Electric field in this region is given by making use of Gauss law in cylindrical symmetry as, $$ \vec{E} = \frac{\lambda}{2\pi r \epsilon_0} \hat{r}$$
Then potential difference is calculated by, $$ - V =  - \int_a^b \vec{E} \cdot \vec{dr} $$ where $\vec{dr} $ is directed from "a" to "b" i.e. radially outwards. So, it becomes $$ V = \frac{\lambda}{2\pi \epsilon_0} \int_a^b \,\frac{1}{r} dr = \frac{\lambda }{2\pi\epsilon_0}\ln{(\frac{b}{a})}$$ 
where again the integral sign canceled by the negative sign(slightly different from the previous - Here we started with finding $-V = V_ - -  V_+$ because the limits will be easy and all are measured from the centre - otherwise $\vec{dl} $ will be directed from -Q to +Q which will have limits 0 to "b-a" where logarithmic function not finite). 
Thus we get the capacitance of the cylindrical capacitor as, $$ C = Q/V = \frac{\lambda{l} 2\pi\epsilon_0}{\lambda \ln{(\frac{b}{a})}} = \frac{2\pi\epsilon_0{l}}{\ln{(\frac{b}{a})}} $$ 

Capacitance of spherical capacitor

Finally, let us consider a spherical capacitor with two concentric spherical metal shells with radii a and b. Inner shell with +Q and outer shell with -Q charge. 
In the same way, the Electric field in the region is given by, $$\vec{E} = \frac{Q}{4\pi\epsilon_0 r^2} \hat{r} $$ vector "r" points radially outwards. 
Potential difference is calculated by, (integration technique is similar to cylindrical capacitor),$$ - V = - \frac{Q}{4\pi\epsilon_0} \int_a^b \frac{1}{r^2} dr $$ Thus,$$ V = \frac{Q}{4\pi\epsilon_0} \left(\frac{1}{a} -\frac{1}{b}\right) $$  
 And the capacitance is given by, $$ C = 4\pi\epsilon_0 \frac{ab}{b-a} $$

Thus we can find the capacitance for various shapes.