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

How to use Gauss's Law to find electric field

W all know that Gauss's law is basically the relation between the charge distribution producing the electrostatic field to the behaviour of electrostatic field in space. Also Gauss's law is based on the fact that flux through any closed surface is a measure of total amount of charge inside that surface and any charge outside that surface would not contribute anything to the total flux. Now we'll go through the main steps which we can employ for applying Gauss's Law

1. First identify the symmetry properties of the charge distribution. By this we mean that the point at which the field is to be determined must lie on a surface and this surface must have enough symmetry which allows integrals involved to be evaluated properly.
2. Determine the direction of the electric field and a surface on which the magnitude of electric field is constant. 
3. Now choose the Gaussian surface accordingly for example if the problem has spherical symmetry then Gaussian surface would usually be spherical and for cylindrical symmetry problem Gaussian surface would be cylindrical.
4. Calculate the flux through the Gaussian surface.
5. Now calculate the charge enclosed inside the chosen Gaussian surface.
6. Equate the two sides of Gauss's law in order  to find the expression for the magnitude of the electric field in that region of space.

How to use Gauss's Law to find electric field

W all know that Gauss's law is basically the relation between the charge distribution producing the electrostatic field to the behaviour of electrostatic field in space. Also Gauss's law is based on the fact that flux through any closed surface is a measure of total amount of charge inside that surface and any charge outside that surface would not contribute anything to the total flux. Now we'll go through the main steps which we can employ for applying Gauss's Law

1. First identify the symmetry properties of the charge distribution. By this we mean that the point at which the field is to be determined must lie on a surface and this surface must have enough symmetry which allows integrals involved to be evaluated properly.
2. Determine the direction of the electric field and a surface on which the magnitude of electric field is constant. 
3. Now choose the Gaussian surface accordingly for example if the problem has spherical symmetry then Gaussian surface would usually be spherical and for cylindrical symmetry problem Gaussian surface would be cylindrical.
4. Calculate the flux through the Gaussian surface.
5. Now calculate the charge enclosed inside the chosen Gaussian surface.
6. Equate the two sides of Gauss's law in order  to find the expression for the magnitude of the electric field in that region of space.

Force on a conductor

We have already learned in our previous discussion that field inside a conductor is zero and the field immidiately outside is
E_n=n(σ/ε₀)                               (1)
where n is the unit normal vector to the surface of the conductor. We also know that any charge a conductor may carry is distributed on the surface of the conductor.
In presence of an electric field this surface charge will experience a force. If we consider a small area element ΔS of the surface of the conductor then force acting on area element is given by
ΔF=(σΔS).E₀       (2)
where σ is the surface charge density of the conductor , (σΔS) is the amount of charge on the area element ΔS and E₀ is the field in the region where charge element (σΔS) is located.
Now there are two fields present  E_σ and E₀ and the resultant field both inside and outside the conductor near area element  ΔS would be equal to the superposition of both the fields  E_σ and E₀ . Figure below shows the directions of both the fields inside and outside the conductor

Now field E₀ has same value both inside and outside the conductor and surface element ΔS suffers discontinuty because of the charge on the surface and this makes field  E_σ on either side pointing away from the surfaceas shown in the figure given above. Since E=0 inside the conductor

E<sub>in=E₀+E_σ=0
E_in=E₀=E_σ
Since direction of  E_σ and E₀ are opposite to each other and outside the conductor near its surface
E_out=E₀+E_σ=2E₀
Thus , E₀ =E/2                      (3)
Equation (2) thus becomes,regardless of the of ΔF=½(σΔS).E                  (4)
From equation 4 , force acting per unit area of the surface of the conductor is
f=½σ.E                                       (5)
Here  is the  E_σ electric field intensity created by charge on area element ΔS at the point very close to this area element. In this region this area element behaves as infinite uniformly charged sheet hence we have,
E_σ=σ/2ε₀                                                  (6)
Now,
E=2E₀=2E_σ=(σ/ε₀)n=E_n
which is in accordance with equation 1. Hence from equation 5
f=σ²/2ε₀ = (ε₀E²/2)n                (7)
This quantity f is known as surface density of force. From equation 7 we can conclude that regardless of the sign of σ and hence direction of E , f is always directed in outward direction of the conductor.

Electric field due to charged conductor

In our previous post we have discussed that electric field inside a conductor is zero and any charge the conductor may carry shall be distributed on the surface of the conductor. For our discussion consider a conductor carrying charge on its surface again consider a small surface element ds over which we can consider surface charge density σ to be approximately constant.
For positive charge distributed over the surface of the conductor , electric field E would be directed at right angels to the surface pointing in outwards direction. Now E due to charge carrying conductor can be calculated using Gauss's law. For this draw a Gaussin cylendrical surface as shown below in the figure

Now S is the area of cross-section of the surface. The flux due to cylendrical surface is zero because electric field and the normal to the surface are perpandicular to each other. Since electric field inside the conductor is zero hence only contribution to the flux is due to the chare on area S lying outside the  surface of the conductor. So total flux through the surface would be

From Gauss's law,
ES=q/ε₀=σS/ε₀
or,
E=σ/ε₀
and this is the required relation for the field of charged conductor

How conductors behave in the presence of electrostatic field

We know that conductors like copper , silver, aluminium etc. , have very large number of free and movable charge carriers , usually one free electron per atom. These free electrons are not bound to its atom and moves freely in the space between the atoms. These free electrons can move under the action of electric field present inside the conductor.
Consider an arbitrary shaped conductor placed inside an electric field such that the field in the conductor is directed from left to right. As a result of this electric field positive charge in the conductor moves from left to right and negative charge moves from right to left. As a result there is a surplus negative charge on the left side of the conductor and a surplus positive charge on the right side of the conductor. This induced surplus chare on both the sides of the conductor acts as a source of an induced electric field which is directed from right to left i.e., in the direction opposite to the initial electric field.
Now with the increase in the amount of induced electric charges, magnitude of induced electric field also increases which cancels out the original electric field having direction opposite to it. This results in a progressive decrease in total field inside the conductor. In the end induced electric field cancels out all the initial electric field thus reaching an electrostatic equilibrium where there is zero electric field at each and every point inside the conductor. Hence we can conclude thet,
E=0 inside the conductor
Now if we apply Gauss's law to any arbitrary surface inside the conductor then total charge enclosed by the gaussian surface equals zero as vector E=0 at all points inside the gaussian surface. From this we conclude that
Al the excess charge (if any) is distributed on the surface of the conductor
We have established the fact that there is no E inside the conductor so tangential component of E is zero on the surface of the conductor hence the potential difference between any two points on the surface of the conductor would also be zero. This indicates that the surface of conductor in electrostatics is equipotential one. Since there is no E inside the conductor so all the points in the conductor are at the same potential.
This is almost all I intended to write in this topic however if you have any doubts then let me know and also you can tell me about the topic you want me to write next in the blog.

Vector Differentiation 1

Here in this post we will revise our concept of Vector Calculas (differentiation of vectors). This mathematical tool would help us in expressing certain basic ideas with a great convenience while studying electrodynamics.

DIFFERENTIATION OF VECTORS
Consider a vector function f(u) such that
f(u) = f_x(u)i + f_y(u)j + f_z(u)k
where f_x, f_y(u) and f_z(u) are scalar functions of u and are components of vector f(u) along x, y, and z directions.If we want to find the derivative of f(u) with respect to u we will have to proceed in the similar manner we used to do with ordinary derivatives thus




where df(u)/du is also a vecor.
Thus in cartesian coordinated derivative of vector f(u) is given by





SCALAR AND VECTOR FIELDS
When we talk about fields then in this case a particular scalar or vector quantity is defined not just at a point in the space but it is defined continously throughout some region in space or maybe the entire region in the space. Now a scalar field φ(x,y,z) assocites a scalar with each point in the region of space under consideration and a vector field f(x,y,z) associates a vector with each point.
In electrodynamics we will come across the cases where variation in scalar and vector fields from one point to is continous and is also differentiable in the particular region of space under consideration.

GRADIENT OF A SCALAR FIELD
Consider a scalar field φ(x,y,z). This function depends on three variables. Now how would we find the derivative of such functions. If we infinitesimal change dx, dy and dz along x, y, and z axis simultaneously then total differential Dφ of function φ(x,y,z) is given as




above expression comes from our previous knowledge of partial differentiation.
If we closely examine above equation this could be a result of dot product of two vectors thus,




or,
dφ=(∇φ)•(dr)
where




is gradient of φ(x,y,z) and gradient of a scalar function is a vector quantity as it is the multiplication of a vector by a scalar.
Thus we see that gradient of any scalar field has both magnitude and direction. Again consider the function φ(x,y,z) then from ordinary calculas any change in this function as discussed above is given by




thus
dφ=(∇φ)•(dr) = |∇φ| |dr|cosθ
FRom this we see that dφ(x,y,z) will be maximum when cosθ=1 which would be the case when dr would be parallel to ∇φ. Thus function φ changes maximally when one moves in the direction same as that of gradient. So we can say that the direction of ∇φ is along the greatest increase of φ and the mahnitude of |∇φ| gives the slope along that direction.
CONCLUSION: The gradient ∇φ points in the direction of the maximum increase of function φ(x,y,z) and the magnitude |∇φ| gives the slope or rate of increase along the maximal direction.

THE OPERATOR ∇
While discussing gradient of a scalar function we find that gradient of any function is given by




or,





where the term in parentheses is called "del"




Del is an vector derivative or vector operator and this operator acts on everything to its right in an expression, until the end of the expression or a closing bracket is reached. There are three ways in which ∇ can act or operate on a scalar or vector function
1. On a scalar function φ : ∇φ (the gradient);
2. On a vector function f, via the dot product: ∇ • v (the divergence);
3. On a vector function f, via the cross product: ∇ x v (the curl).
Out of these three ways of operation of ∇ on any function we have already discussed gradient of a scalar function.

In this post we learned about scalar and vector fields, gradient of scalar fields and ∇ operator. In the next post we'll lern more about vector differential calculas i.e, in particular we'll discuss divergance and curl of vector fields.
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