Electrostatics(part-1)

Electrostatics(part-1)

Electrostatics:- The Electrostatics deals with all the Phenomena arising out of system of static charges. We are into the domain of physics where interaction takes place between two bodies at a distance. There are two kinds of interaction between two bodies:-

One involves the direct contact of two bodies.

Other does not involve direct contact of two bodies.

It look very surprising how two bodies can interact with each other without being in contact. How do two human beings interact? They interact through exchange of ideas, thought or by looking at each other. All kinds of interaction between human beings involve exchange of something. You cannot imagine interaction between two blind & deaf & dumb individuals if they remain at a distance. This gives a kind of intuition that interaction between two charges might involve exchange of some particles.

Our intuition turns out to be correct. Yes! The interaction between two charges does involve exchange of a massless & chargeless particle called photon.

We can well appreciate the concept by taking an example. Consider two guys who are standing on a movable platforms which lies on a frictionless floor.

When they exchange balls they tend to move backward as if they repel each other. So, the exchange of ball has the effect same as repulsion.

Now consider same guys are snatching balls from each other. This kind of exchange of ball leads to the effect same as attraction.

So, the exchange of particles between two bodies leads to some kind of force getting generated between two bodies. So, interaction at a distance always involves an exchange of particles. In our case, the electrostatic force between two charges involves continuous exchange of particles. Further discussion on this topic is beyond the scope of this article.

Now we will discuss various aspects of electrostatic interaction.

1. Quantisation of charge:- The magnitude of charge in any electrostatic system must be multiple of electronic charge.
   

   q = ne
   

   where n is integer.
   

   E is the electronic charge e = 1.6 x 10^-19C
   

2. Coulomb’s Law
   

According to coulomb’s law the force between two charges is equal to

q₁ and q₂ are the magnitude of two charges.

r is the distance between two charges.

is permittivity of space

So, magnitude of force between two charges q₁ & q₂ is

The direction of force is along the line joining two charges. This kind of force which act along the line joining two bodies is called central force. Gravitational force is a similar kind of force.

At this point, we must introduce the concept of electrostatic field or electric filed E. Before directly moving to numerical formulation of electric field, let us first discuss electric field itself . What is field?

What happens if great film star Amitabh Bacchan comes to your locality. You may run to take his autograph. He creates a field around himself and when you happen to come within the limits of his field, you get attracted. If he is in Mumbai and you are in Delhi, it is unlikely that you may not move to Mumbai for his autograph. This means that his field decreases with the distance.

We have two concept involved in field. What is cause of field and how field varies with distance. Strength of field created by any person depends on his/her personality. The magnitude of field always varies inversely with distance.

Similar is the case with field produced in the neighbourhood of charge. If any other charge comes in the vicinity of this charge it gets attracted or repelled in the presence of its electric filed depending on the nature of charges.

Now let us move to mathematical formulation of electric field. The strength of electric field produced by any charge depends on the magnitude of charge itself and the magnitude of electric filed depends inversely on square of its distance.

So, electric field due to a point charge q at any point P is defined by force acting on unit test positive charge at point P.

Where q₀ is unit test prositive charge.

Usually, we will come across the problem which involve the calculation of electric field due to

1. System of descrete charges
   
2. System of continuous distribution of charges
   

System of discrete point charges

Electric filed due to system of discrete point charges is calculated by superposition of electric field due to individual charges.

The resultant electric field

Where E_i is the electric filed due to ith charge.

Superposition of Principle

Before discussing superposition principle, let us discuss a very obvious but very important concept that is principle of cause & effect. What is this principle? When you see smoke coming out, you immediately conclude that there must be fire which is producing smoke. If a mass accelerates, some unbalanced force must be acting on it. If heat is flowing along strip of metal, there must be temperature difference at the two ends of strip. If a electric current is flowing in a conductor, there must be potential difference applied across it.

It means where ever you see some effect there must be some cause present.

Now a quite obvious question comes to our mind. That is how effect depends on cause. In most of the physical system but not all, effect depends linearly on cause. Such a system is called linear system. Electromagnetic system is also linear system.

In linear system effect Cause

So, our electromagnetic system ,mechanical system, thermal systems are all linear system.

Here we will introduce a very important concept of inertia of a system. Inertia of any linear system is basically the measure of tendency of system to minimize effect produced by the cause or it is tendency to oppose any change of state.

So, we can write

Cause = inertia x effect

For example

Force = mass x acceleration

Torque = MI x angular acceleration

Now, let us consider a case, where two causes C₁ & C₂ when act individually produce an effect E₁ & E₂.

C₁ = Inertia x E₁ ……….(1)

C₂ = Inertia x E₂ ……….(2)

When two causes act simultaneously on the linear system

Then Cause    C = C₁ + C₂

= Inertia x (E₁ + E₂)

= Inertia x E

Where E = E₁ + E₂

Here we see that, when two causes act simultaneously on any system, the net effect is the addition of their individual effect.

Let us take one example

Let F₁ & F₂ be the two force when acting individually on mass m, produce individual acceleration a₁ & a₂

When two force act simultaneously on the mass m the net acceleration is the sum of their individual accelerations

This is called superposition principle and it is applicable to all kind of linear system.

Exercises

1. Two particles A and B having charges 8.0 x 10^-6C and -2.0 x 10^-6C respectively are held fixed with a separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force?
   
2. Three charges, each equal to q, are placed at the three corners of a square of side a. Find the electric filed at the fourth corner.
   
3. Four particles, each having a charge q, are placed on the four vertices of a regular pentagon. The distance of each corner from the centre is a. Find the electric field at the centre of the pentagon.
   
4. Find the electric field at a point P on the perpendicular bisector a uniformly charged rod. The length of the rod is L, the charge on it is Q and the distance of P from the centre of the rod is a.
   

   The electric field is defined for point charge. But by incorporating superposition principle, we can find electric field due to distributed charge system also. Now, we will see few cases of distributed charge system.
   

Electric field due to continuous change distribution we have seen how to find electric field due to system of discrete print changes let us now me how we can find electric field due to anttinuous change distribution we have there kind of change distribution.

1. Linear change distribution–Change is distributed linearly. For example–Charge is distributed over a rod of length l we define a term linear change density  such that
   

   
   

If charge is uniformly distributed

Consider a change Q distributed over a length L Suppose if we want to find electric field at a print P at a distance d above the mid point of rod.

Here also we apply the principle of superposition. Here we consider a small element dx at a distance x from mid point of rod.

Change on dx = dq =

Electric field at a print P due to

So, the net electric field at point P will the superposition of field due to all the elements.

The horizontal components of field due to individual element add upto zero.

Vertical component of dE

x

So net field

Put x = d tan     

Uniformly charged ring

Suppose a charge Q is uniformly distributed over a ring and we have to find electric field at any axial print P

Linear change density

Now, we take an element ds on the ring do change dq on ds = ds

Electric field dE

Now, if we super impose the field due to all such element at point P. then the vertical component of dE will cancel and any axial component of E will get added up producing resultant is that direction

‘So, net field E=

in the direction of OP

(2) Surface charge distribution. When charge Q is distributed over a surface we say charge has got a surface distribution

We define a term surface charge density  such that

If charge is uniformly distributed

or     Where S is surface area

For example suppose charge Q is uniformly distributed over a disc of radius R and we have to find electric field at any axial point P.

Let us consider a ring of radius r and thicknes dr

Charge on ring =

Now, electric field at print P

…(i)

1. Net field at point P will be calculated by superimposing field due to all such element ring. This can be obtained by integrating (1) Over the surface of disc
   

Put

toward

Electrostatic Potential

Electrostatic potential at a point P is the work done in moving a test unit positive charge from some reference point to point P such that charge is never accelerated in such a process i.e. it should always remain in quasistatic equilibrium.

If charge is accelerated in such a process then work done will be equal to potential energy+ kinetic energy.

Reference point is taken such that there is no interaction between charges at that point.

Usually infinity is taken as reference point

So, The unit of potential is

in MKS system

or

Volt in SI system

Force applied while doing the work should always be equal and opposite to the electrostatic force acting on the unit positive charge or electric field at that point.

Work done in displacing a unit positive charge by a distance

so potential at point P is

Potential is a point function.

Work done in displacing a test positive charge from point A to point B is difference between potential at these point and since potential is point function, it is independent of path along which it is displaced.

If charge is displaced in a loop

work done is zero.

Calculation of Electrostatic potential

If reference point is taken at infinity where the potential is assumed to be zero and electric field is assumed to be due to a point charge.

i.e.

i.e.

For a system of discrete charges, potential at a point is calculated by using principal of superposition.

Where is the distance of a point from the charge

If there is continuous distribution of charge within a finite volume.

(1) Find the electric potential due to conducting sphere of radius R having a charge Q .

Solution :

Let us find potential at point P

work done to bring charge from infinity to point P

For point P inside sphere

Electric field inside sphere E = 0

Electric potential inside conducting sphere is constant and equal to the value of potential at the surface of conductor.

(8)

Figure shows three concentric thin spherical shells A, B and C of radii a, b and c respectively. The shells A and C are given charges q and –q respectively and the shell B is earthed. Find the charges appearing on the surfaces of B and C.

Solution :

Conducting sphere is equivalent to two non-conducting surfaces.

Here we will make use of Gauss Law and the fact that Electric field inside conductor is zero.

Now, let us take Gaussian surface between 1 & 2.

E = 0 at each point of Gaussian surface.

So, it should enclose no charge.

So, charge on inner surface of A is zero.

Hence, outer surface of A will have q amount of charge.

Surface B is earthed and earth is unlimited source and unlimited sink of charge.

Surface B will have that amount of charge so that its potential is zero.

Let us take Gaussian surface between 3 & 4

Electric field E = 0

So, it will enclose zero charge.

So, inner surface of B will have –q charge

Let outer surface of B has q’ charge

Similarly, inner surface of C will have – q’ charge

Outer surface of C will have (– q + q’) charge.

Now, potential of surface B is zero.

Applying principle of superposition, Potential at B will be sum of potential due to all the six surfaces independently.

So, charges on

surface 1 = 0

surface 2 = q

surface 3 = –q

surface 4 =

surface 5 =

surface 6 =

Potential due to dipole

Two equal point charges q of opposite sign separated by a small distance 2l constitute an electric dipole of electric dipole moment. It is denoted by p.

the direction of (2l) being from negative to the positive charge.

charges are separated by a very small distance.

An example of dipole is hetro-nuclear diatomic molecule. Due to difference in electro negativities, the atoms tends to develop partial positive and partial negative charges which are at a distance equal to bond length.

HCl molecule acts as a dipole

The potential at P due to the dipole will be

Dipole in an Electric Field

Net force on the dipole is zero.

Torque on the dipole is given by

Thus, an electric dipole placed in an electric field experience, a torque which tends to align it with the field.

In order to change the orientation of an electric dipole in an external field, the external agent must do certain work. This work is stored as potential energy in the system.

Now, work done in rotating the dipole through an additional angle

Therefore, the work done in rotating the dipole from an initial position to final position

This work done is stored up in the form of potential energy of the dipole in the final position.

We define the potential energy of a dipole making an angle 90° with the electric field as zero.

Where U is potential energy of dipole.

p is dipole moment

E is electric field.

Dipole-dipole interaction energy

Dipole-dipole interaction energy

Energy of interaction of dipole in the field of another dipole is called dipole-dipole interaction energy.

If there are two dipoles then one dipole interacts with the electric field produced by another dipole.

Energy of interaction

where E1 & E2 are electric field due to first & second dipole respectively.

We will cover Gauss Law in the next article.

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