GENERALIZATION OF FARADAYS EXPERIMENTS WITH CONCENTRIC SPHERES

The results of Faraday experiments with concentric spheres can be summed up as an experimental law by stating that the electric flux passing through any imaginary sphere placed between two conducting sphere is equal to the charge enclosed within the surface of imaginary sphere. This enclosed charge is distributed either on the surface of the imaginary sphere or is located at a point at the center of the imaginary sphere.


However since 1 C of charge produces 1 C of electric flux, the inner conductor might just well have been a cube or a pot shaped conductor, but still the charge on the outer sphere will be the same. Of Course the flux density would change from its symmetrical distribution to an unknown configuration, but +Q C on any inner conductor will induced a –Q C of charge on surrounding sphere. Going a step further replacing outer sphere with a cylindrical conductor and inner sphere with a pot shaped conductor with charge Q. Producing Flux = Q C the induced charge in the cylindrical conductor will be –Q C. How Fascinating right??


This generalization lead to many developments one of them is discussed later in the next post. All the experiments performed by faraday on electromagnetism is a must read for every engineer. He gave us modern electromagnetics which is used almost in every hardware. I would try to post all his experiments, conclusions and generalization details on the site but self reading is the best reading.


The post here under category Electromagnetics is taken from my college blog. They are updated here first by me then brought to this site again by me.

Electric Field Intensity

Consider a fixed charge Q1. Let us move a second charge say test charge Qt around the Q1. We observe that the test charge Qt experiences force everywhere around Q1. It is experiencing a force field. Force on it will be given by coulombs law as:

But still question arises what is Electric Field?? Before defining electric field let us modify above equation a bit.
Let us write the above equation as force/charge equation:

This quantity that is force/charge describes a vector quantity. This vector quantity is called Electric Field Intensity.So Electric field intesity is defined as the vector force on an unit positive test charge by a fixed charge Q1. Or it can also be defined as force per unit charge.
So to sum it up we can Electric field intensity as:

So it becomes.

Expressing Electric field in cartesian Coordinate:

Expressing Electric Field in Spderical Co-ordinates:
First for considering spherical co ordinate we must see that fixed charge has spherical symmetry at its location like at origin(0,0,0) for that we can either use spherical co-ordinate directly in equation as:

Where Er and r spherical co-ordinates values or we can convert them in cartesian coordinate which will give us the equation as below:
We will have r = R = xi + yj +zk ar = aR = (xi + yj + zk)/√(x^2+y^2+z^2 )

If more charges are added in the surrounding of Q1 then electric field will be sum of individual acting alone. As shown in equation given below:



Before ending the topic let us do a small quick example.
Example: See the figure below. We need to find out Electric field at P(1,1,1) caused by 4 identical 3nC charges located at P1(1,1,0), P2(-1,1,0), P3(-1,-1,0) and P4(1,-1,0).

Example of Electric field

[caption id="" align="alignnone" width="615" caption="Example of Electric Field"][/caption]


Answer: As r = i + j + k and r1 = i+j therefore r – r1 = k. Therefore |r – r1| = 1.
Similaraly r – r2 = 2i + k. therefore |r – r2| = √5
Similaraly r – r3 = 2i + 2j + k. therefore |r – r3| = 3
Similaraly r – r4 = 2j + k. therefore |r – r4| = √5
And as




Therefore simplifying we get E =
E = 6.82i + 6.83j + 32.8k V/m ans.

The Coulombs Law

Columbs law state that the force between two small objects seperated in a vaccum by distance much greater that the size of each object is proportional to the product of charges on them and inversely proportional to square of distance between them.

where Q1 and Q2 are positive or negative charge quantity measured in columbs and R is the distance between them measured in meters. Here k is proportionality constant given as:

Again another constant € is called permittivity of free space. Measured in Farad per meter(F/m).

The above quantity is not dimensionless. For coulombs law dimensions are C2/N.m2 . This means Farad has units C2/N.m .This makes are coulombs law as:



VECTOR FORM OF COULOMBS LAW:


[caption id="" align="alignnone" width="349" caption="Vector form of Coulombs law"][/caption]


Let us say charges Q1 and Q2 are like charges distance of each from origin be r1 and r2 respectively. As F2 is in direction of R12. The vector R12 is cleary
R12 = r2 – r1.
Therefore the vector form of coulombs law is:

Where a12 is a unit vector in direction of R12 or we can say:



Let us finish this topic with an example.
Example: let umm. There be 2 charges say Q1 = 3 x 10^(-4) C located at say M(1,2,3) and Q2 = 10^(-4) C located at say L(2,0,5). We require to find force exerted by Q2 on Q1.
Answer: Let F2 = force exerted by Q2 on Q1.
Vector R12 = r2 – r1 = (2 - 1)i + (0 – 2)j + (5 – 3)k = i – 2j + 2k
This gives us |R12| = 3
Which again gives as unit vector

Therefore our F2



F2 = 10i – 20j + 20k N ans.
Therefore we found out F2 that is force of Q2 on Q1. Similary we can find F1 that is force of Q1 on Q2. Interestingly as both charges are like charges. We will find out that



Coulombs law is linear. We can see that multiplying Q1 by some factor n. We find that the force between the 2 charges also get multiplied by factor n.
Another thing is that the force on the charge in presence of the other charges is sum of forces on that charge due to each of the other charges acting alone.