Question

\(\huge\color{blue}{\text{Conceptual Guide to Electrostatics.}}\)

Answer 1

The study of stationary electric charges or fields is called as \(\bf Electrostatics.\)

Answer 2

\(\huge\bigstar\rm Coloumb's~Law\)

\(\boxed{\huge\ F=k\frac{q_1.q_2}{r^2}}\)

Where,
\(\bigstar\) k is called as \(\bf Electric~Force~Constant.\) It is equal to \(\huge \frac{1}{4\pi\epsilon_o}\)
\(\bigstar\) \(\epsilon_o\) is \(\bf Electric~permittivity~of~free~space\) and in SI unit its value is equal to
\(\bf 8.854*10^{-12}\)\(C^2N^{-1}m^{-2}\)
\(\bigstar\) Plugging in these values we get the value of k as \(\bf 9*10^9C^{-2}Nm^2\)

Answer 3

\(\bigstar\) \(\bf Dielectric~constant~or~Relative~electrical~permittivity\)
It is the ratio of absolute electrical permittivity of free space. It can also be defined as the ratio of force of attraction between two charged particles in the free space to the force of attraction in the given medium.

\(\boxed{\huge \ K=\frac{\epsilon}{\epsilon_o}=\frac{F_o}{F}}\)

Answer 4

\(\huge\color{green}{\text{Superposition Principle}}\)
According to it,
Total force on any charge due to a number of other charges at rest is the vector sum of all the forces acting on it by the other charges, taken one at a time. The force due to individual charges are unaffected due to the presence of other charges.

Answer 5

For example,
\(\color{red}{\rm Three~equal~charges,~2*10^{-6}C~each,~are~held~fixed~at~the~corners~of~an~equilateral~triangle~\\of~side~5cm.~Find~the~coulomb~force~experienced~by~one~of~the~charges~due~to~the~other~two~\\charges}\)

Answer 6

\(\color{blue}{\bf Solution:}\)
Let each charge be represented by q and side of the triangle be r, then force on C by A and B individually will be equal to \(\huge\frac{kq^2}{r^2}\). The net force will be equal to the vector addition of these forces at 60°.

\(\huge F_{net}=2*\frac{kq^2}{r^2}*Cos30°=\sqrt{3}\frac{kq^2}{r^2}\)

Plugging in the values we get 24.94\(\approx\)25N

Answer 7

\(\bigstar\bf Electric~Field\)
It is the space around a charge within which a force would be exerted on other charged particle.

\(\bigstar\bf Electric~Field~Intensity\)
It is the strength of electric field at any point. For any point it can be defined as the force experienced by a unit positive charge placed at that point.

\(\boxed{\huge\overrightarrow{\rm E_r}=\frac{\overrightarrow{\rm F_r}}{q_o}}\)

\(\bigstar\) Where, \(F_r\) & \(E_r\) are the force and Electric field intensity at position r.

Answer 8

\(\bigstar\) Electric field intensity E\(_r\) at a distance r due to a point charge q can be expressed as following:

\(\boxed{\huge\rm |\overrightarrow{E_r}|=\frac{q}{4\pi\epsilon_or^2}}\)

\(\bigstar\) The electrostatic force between the charge Q and q can be looked upon as an interaction between charge q and electric field of Q and vice-versa.
\(\bigstar\) For a +ve charge, the electric field will be directed radially outwards from the charge. On the other hand, if source charge will be negative, the electric field vector will be inwards.
\(\bigstar\) The electric field intensity has spherical symmetry.
\(\bigstar\) Electric field intensity also follows superposition principle.

Answer 9

\(\huge\text{Physical Significance of Electric Field}\)
Suppose we consider two charges q\(_1\) and q\(_2\) in accelerated motion. Now any signal or information which can go from one point to another is C (speed of light). Thus, effect of any motion of q\(_1\) on q\(_2\) cannot arise instantaneously. There will be some time delay between the effect (force on q\(_2)\) and the cause (motion on q\(_1\)). Now according to field picture the accelerated motion of charge q\(_1\) produces electromagnetic waves, which then propagates with the speed C, reach q\(_2\) and cause a force on q\(_2\). The notion of field elegantly accounts for the time delay. Hence, even though electric and magnetic fields can be detected only by their effects, they are regarded as physical entities.

Answer 10

With this knowledge, we could find the characteristic impedance of free space.

Answer 11

Simpler SI unit for \(\large \epsilon_o\) is F.m\(^{-1}\)

Answer 12

\(\huge\text{Gauss's Law}\)
The law states that net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface. quantitatively,

\(\Large \boxed{\large \phi=\frac{q}{\epsilon_o}}\)

Where, \(\phi\)=Electric flux=\(\large \int_{}^{}E.dA\), q=net charge inside the body, A=surface area of the body, \(\epsilon_o\)=permittivity of free space.

Answer 13

Let's try to see if it's valid or not !

Suppose we take a sphere with radius r and small area element \(\triangle\)S

Answer 14

If q is the net charge contained within it then,
Electric flux = \(\int E.\triangle S\)=\(\huge{\int\frac{q}{4\pi\epsilon_or^2}}.\small \triangle S\)

Answer 15

= \(\huge \frac{q}{\cancel{4\pi r^2}\epsilon_o} \times\large \cancel{4 \pi r^2}\) = \(\huge\frac{q}{\epsilon_o}\)
Hence, the law is valid.

Answer 16

More!

Answer 17

Electric Field Potential, Torque due to Electric Field, Conservation of Energy.

Answer 18

@Johnbc Keep Clam !
Coming up :)

Answer 19

Great work @Abhisar.

Though, if I may suggest, please consider using vectorial notations more frequently as students generally get confused when they are told that area is a vector quantity. Such notations will also help them in understanding that why in some cases flux may be zero even when electric field and area are non zero.

But, fantastic work. Even sparing time for creating such a great guide is appreciable.

Vishwesh

Answer 20

Yep, Thanx for the feedback i'll keep that in mind for rest of the part.

Answer 21

No problem :)

Best wishes !

Answer 22

U too !

Answer 23

This is great, nicely done @Abhisar

It's hard to keep up with all your tutorials, just amazing.

Answer 24

Master piece. excellent

Answer 25

Thnx @astrophysics and @no.name !