Complete notes of electrostatics –

 

Here complete notes on electrostatics is provided by physics classes by Nayan jha . This notes are based on CBSE  Delhi syllabus . This is the combination of all the topics already provided on this website , which can be studied by the students one by one ( individually )  . If students wants to see all the notes on single page  can go through this page . The link of the all topics are given below –

Topic-1 -; What is electric charge ? And  what are the properties of charge? 

Topic-2-; Coulomb’s law of electrostatics (Vector form)

Topic-3-;  Electric field , and electric field lines . All the properties of electric field lines  

Topic-4-; Electric dipole , electric field at a point on axial and equatorial point due to dipole .

Topic-5-;Electric field at any point due to a dipole 

Topic-6-; Expression for torque and potential energy , when a dipole is placed in external uniform electric field.

Topic-7-; Electrostatic potential and potential energy

Topic-8-; Equi-potential surfaces

Topic-9-; Electric flux and area vector

Topic-10-; Gauss’s law , proof of Gauss’s law , and prove of Coulomb’s law with the help of Gauss’s law

Topic-11-; What do you mean by the term electrical capacitance?

Topic-12-; Explain  parallel plate capacitor , find expression for capacitance of parallel plate capacitor .

Topic-13-; Grouping (combination ) of capacitors ( series and parallel combination)

Topic-14-; Energy stored in capacitor and energy density

Topic-15-;Common potential and loss in energy on shearig charges

Topic-16-; Non-polar and polar dielectrics

Topic-17-; Capacitance of parallel plate capacitor with conducting and dielectric slab 

 

 

 

Capacitance of parallel plate capacitor with conducting and dielectric slab.

Capacitance of a parallel plate capacitor with a conducting slab –

Let A is the area of plates and d is the separation between them is d then capacitance C₀= ϵ₀A/d .

Let Q is the charge on the plates of the capacitor. When a conducting slab of thickness ‘t’  ( t< d ) is inserted between the plates . Then the original electric field E₀ exist over a distance (d-t) , And inside the metallic plate electric field is zero . so net potential difference between the plates is

V = E₀ (d – t) = σ(d-t)/ϵ₀ ;   where σ=Q/A

As C = Q/V    = Q /( σ(d-t)/ϵ₀ ) ;  But , σ = Q/A

C = Aϵ₀ /(d-t) = Aϵ₀/[d (1-t/d)] = C₀/(1-t/d)

Special case if    t = d  then   new capacitance of the arrangement will be  infinity ,

Capacitance of a parallel plate capacitor with a dielectric slab  –

Let A is the area of plates and d is the separation between them is d then capacitance C₀= ϵ₀A/d .

Let Q is the charge on the plates of the capacitor. When a dielectric slab of thickness ‘t’  ( t< d ) , and dielectric constant K  is inserted between the plates . Then the original electric field E₀ exist over a distance (d-t) , And inside the dielectric plate electric field is E₀/K . So net potential difference between the plates is ,

V = E₀ (d – t) +( E₀ t/K)  = E₀[d-t+(t/K)]

= σ[d-t+(t/K)] /ϵ₀  ,   where σ=Q/A

As C = Q/V    = Q / σ[d-t+(t/K)] /ϵ₀  ;  But , σ = Q/A

C = Aϵ₀ /(d-t +t/K)= Aϵ₀/[d (1-t/d+t/dK)] = C₀/(1-t/d  +t/dK) ;

If t= d   then  new capacitance  C = KC₀

 

*** Special cases –

When a dielectric  slab of dielectric constant ( K ) is inserted between the plates having same thickness of separation between the plates . (t = d ) . 

 

When battery disconnected             

( I ) Capacitance C = KC₀

(II) Potential  V =  V₀/K

(III) Charge  Q = Q₀

(iv) potential energy U = U₀/K

  When battery remain connected       

(I) Capacitance C = K C₀

(II) Potential  V = V₀

(III)  Charge   Q = K Q₀

(IV)  potential energy  U = KU₀

 

Non-polar and polar dielectrics –

In this topic we will discuss about Non-polar and polar dielectrics and dielectric polarization with electrical susceptibility.

Non polar and polar dielectrics  –

As we know the dielectrics are insulating material which transmit electric effect without conducting electricity .

Dielectrics is of two types , Non-polar Dielectrics and polar Dielectrics .

In non-polar dielectric ,  in the molecule of the dielectric the centre of mass of the positive charge coincide with the centre of mass of the negative charge . Like , Hydrogen , nitrogen , oxygen , carbon dioxide , methane , benzene etc . In such kind of molecule each molecule has zero dipole moment and they are symmetric in shape .

But in case of polar dielectric . The  centre of mass of positive and negative charges do not coincide  because of the asymmetric shape of the molecules. It has permanent dipole moment . example – HCl , NH₃ H₂O, alcohol etc .

Dielectric polarization –

When a non polar dielectric is held in external electric field , the two centres of positive and negative charges in the molecules are separated . The molecules gets distorted and is said to be polarized as a small dipole moment .

The dipole moment acquired is given as , p = αϵ₀E₀  Where α is the constant of proportionality called molecular polarizability .  Its unit is m³ .

If N is the number of atoms per unit volume then , electric polarization P = N p   or,  P = N qx .

Its unit is C-m^-2

If we consider a small volume element in the interior of the slab  as shown in figure . , each volume has no net charge ,though it has net dipole moment . This is because positive charge of each dipole sits close to the negative charge of the adjacent dipole . . The negative end of the dipole remain un-neutralized  at the surface AB and positive end of the dipole at CD . then the effective electric field E = E₀ -E_P  ;

Further E_P = σ/ϵ =  p/ϵ.

Electric susceptibility –  As we know that the electric polarization P αE    or, p= ϵ  χ E

Where  χ is constant called electrical susceptibility .

For vacuum χ = 0 ;

But as we know ,   E = E₀ -E_P  = E₀ – P/ϵ = E₀  –  ϵ  χ E/ϵ

Or , E = E₀ -χE  ;

E₀ = E  +   χ E  = E( 1+ χ )

OR ,             E₀/E  =  1 + χ

K ( Dielectric constant or relative permittivity ) = 1 + χ   [ since  K= E₀/E  ]

K (ϵ_r) = 1 +  χ …………………This is the relation between relative permittivity and susceptibility .

Common potential and loss in energy on shearing charges .

Before to know about the topic Common potential and loss in energy on shearing charges .student must have to know about Energy stored in the capacitor and electrostatic potential .

To know about energy stored in a capacitor click here-

To know about the electrostatic potential energy click here- 

In this topic students will know about Common potential and loss in energy on shearing charges . its derivation and numerical is important for class 12th ( Board examination) and for NEET/JEE examinations .

Common potential – When two charged capacitors are connected with each other using a conducting wire then charge starts flowing from high potential to low potential, and after some time their potential becomes equal.

Let two capacitors C₁ and C₂ and their potentials are V₁ And V₂ respectively. Then total charge before combination Q = C₁V₁ + C₂V₂.

Let they are connected with a conducting wire, then charge start flowing from high potential to low potential, till potential is not equal. Let V is the common potential then total charge after combination Q = V (C₁ + C₂).

But from the conservation of energy charge after collision = charge before collision

So, C₁V₁ + C₂V₂ = (C₁ + C₂) V

Then     V = C₁V₁ + C₂V₂ / (C₁ + C₂)………………………eq.

 

Loss of energy on shearing charges –

As we have seen in the above equation  common potential after combination

V = C₁V₁ + C₂V₂ / (C₁ + C₂)

Total energy before combination U₁ = ½ C₁V₁² + ½ C₂V₂²

Total energy after combination  U₂= ½ ( C₁ + C₂ ) V² ; Putting the value of  V

We get , U₂= ½ ( C₁ + C₂ ) [C₁V₁ + C₂V₂ / (C₁ + C₂)]²

= ( C₁V₁+C₂V₂ )² /2(C₁+C₂ )

Then loss in energy ;

=U₁-U₂ = [½ C₁V₁² + ½ C₂V₂²] – [( C₁V₁+C₂V₂ )² /2(C₁+C₂ )]

= [C₁V₁²(C₁+C₂ )+C₂V₂² (C₁+C₂ )- (C₁V₁+C₂V₂)²]/2(C₁+C₂) ;

= C₁C₂(V₁²+V₂²-2V₁V₂)/2(C₁+C₂) Or  U₁-U₂ = C₁C₂(V₁-V₂)²/2(C₁+C₂) . Which is positive for all  possible values of  C₁ , C₂ And V₁ , V₂ .

I.e .  U₁ – U₂ >0 ( POSITIVE ) .

So, U₁ > U₂ . Therefore we can say that there is always loss in energy during the combination of charged capacitors .

 

Grouping of capacitors

Before to know about grouping of capacitors students should know about parallel plate capacitor .To read about the topic parallel plate capacitor click here-

This is the topic in which students will learn about grouping of capacitors , which contain series combination and parallel combination of the capacitors .

(Combination) Grouping of capacitors or condensers  –

1. Capacitors in series –   When negative plate of first capacitor is connected with positive of second , negative of second connected to positive of third and so on such combination is called series combination . in another words we can say that, the capacitors are said to be connected in series between two points , when we can proceed from one point to the other only through one path .

As shown in figure three condensers of capacitance C₁ , C₂ and  C₃  are connected in series .

Suppose V is the potential difference across the combination .  Suppose   ‘Q’ charge is given to the left plate of C₁ due to induction -Q charge setup on the other plates and =Q on the left plate of capacitor C₂ and so on . Let V₁, V₂ and V₃ are the potential difference setup across the capacitors  C₁ , C₂ and  C₃  respectively . charge across the capacitors are same which is Q . Then V₁ = Q/C₁ , V₂ = Q/C₂ . V₃ = Q/C₃ . So net potential   V = V₁ + V₂ + V₃ ;

If C_S is the equivalent capacitance of the combination of the capacitors then  ,

Q/C_S = Q/C₁ +  Q/C₂ + Q/C₃

Or ,  1/C_S = 1/C₁+ 1/C₂ + 1/C₃

If there are n  capacitors are connected in series then

 1/C_S = 1/C₁+ 1/C₂ + 1/C₃ …………………+ 1/C_n

 

2. Capacitors in parallel – The combination of the two or more capacitors are said to be in parallel if positive plates of capacitors are connected at one point and all negative plates at another points as shown in figure .

Let C₁ ,C₂ , and C ₃ are the capacitance of the three capacitors , and V is the potential difference applied across the combination . The potential difference across each capacitor is V the charge on each are Q₁ Q₂ and  Q₃ , because their capacitance are different .  S o net charge Q = Q₁ + Q₂ + Q ₃

If C_P is the equivalent capacitance of the capacitors then C_P V =  C₁ V + C₂ V + C₃ V

So, Equivalent capacitance  C_P  =  C₁  + C₂  + C₃ ;

If there are n  capacitors are connected in parallel then

C_P  =  C₁  + C₂  + C₃……………..+ C_n

 

 

Parallel plate capacitor

Before to know about these topics student must know about Electrical capacitance, To know about electrical capacitance click here-

In this topic we will discuss about capacitance and its principle ,and expression for the capacitance of parallel plate capacitor .

Capacitor and its principle – It is an arrangement in which large amount of  electric charge can be stored and hence develop large potential difference in small space. A capacitor consists of a system of two conductors separated by an insulating medium. Often the two conductors are charged by connecting them to the two terminals of a battery.  Or we can say that the capacitance of an insulated conductor is increased considerably by bringing near it an uncharged conductor connected to earth.

 If the conductors are plane , the capacitor is called a parallel plate capacitor , when the conductor is spherical called spherical conductor , and if the conductor is cylindrical then the capacitor is called cylindrical capacitor .

Parallel plate capacitor – A parallel plate capacitor consists of two thin conducting plates each of plate area ‘A’  and separated by distance ‘d’ . One of the plate is insulated and the other plate is earthed .  When charge Q is given to one plate then the charge -Q is induced on the near face of the  second plate and +Q on the farther  face of the second plate , and +Q is flows to the earth .

The surface charge density of the first plate is     +σ = Q/A

And on the other plate is        -σ = -Q/A .

So electric field between the plates is  E = σ/ϵ₀ = Q/Aϵ₀ ;

Potential difference between the plates is V = E.d = Qd/Aϵ₀ 

Hence capacity of the  capacitance will be C = Q/V = Q/( Qd/Aϵ₀ ) = Aϵ₀  / d   i.e . the capacitance of parallel plate capacitor is independent of the charge and potential .

 

Electrical capacitance

In this topic we will discuss about the Electrical capacitance and electrostatics of conductor,electrostatic shielding  , and the capacity of an isolated spherical conductor .

Electrostatics of conductor –

As we know a conductor is a substance through which electric charge can conduct from one point to another . But there is some important results regarding  electrostatics of conductor which are follows –

1. In electrostatic equilibrium , electric field is zero every where inside the conductor . Suppose a conductor ABCD is held in an external electric field of intensity E₀ as shown in figure , the free electrons in the conductor moves from AB to CD . As a result , some net negative charge appears on AB . and the produced electric field E_p , which opposes the flow of free electrons from AB to CD . since Ep = E₀ so net field inside the conductor is zero .

2. The electric field just outside the charged conductor is perpendicular to the surface of the conductor at every point .
3. Net charge in the interior of the conductor is zero .

According to Gauss’s law  ∫E.ds =q/ϵ₀ ];

since inside the conductor E=0

therefore q= 0 also ;

4. Electric field at the surface of the charged conductor is given as E = σ/ϵ₀ where σ is the surface charge density . and the surface charge density is different at different point for irregular shaped conductor .

ELECTROSTATIC SHIELDING –

 It is the phenomenon to protect a space or an object from external electric field by covering it by conducting substance . as we know electric field inside a conductor is zero hence when an object is placed inside the conductor then there is no electric field connected to the object which is inside the conductor .

Electrical capacitance –

Electrical capacitance of a conductor is the ability of the conductor to store electric charge in it.

When conductor given some charge then its potential increases . Let Q is the charge given to the conductor then its potential increases to V volt so , we can say,   Q α V  or , Q = CV ; where C is a constant called capacity of the conductor , we can write  capacity C = Q/V  , Using this expression we can write , capacity of a conductor is the ratio of charge given to the potential developed in the conductor i.e. C = Q/V ;

 

In another way we can define , electrical capacity of a conductor is numerically equal to the charge required to increase the potential by unit volt .      Its unit is farad (F)

Here 1 Farad =1 coulomb/1 volt and its dimension is [ M^-1L^-2T⁴A²]

CAPACITY OF AN ISOLATED SPHERICAL CONDUCTOR –

Suppose a conductor of radius ‘r’ with centre ‘O’ has given charge ‘Q’  then its potential increases by ‘V’ . here potential V= q/4∏ϵ₀r

As we know capacity C = Q/V = Q/( q/4∏ϵ₀r) Or. C = 4∏ϵ₀r  ; Here r is in meter and  C is in farad .

As we can see C α r i.e larger the radius of the sphere and larger will be capacity of the conducting sphere .

Using this formula we can find the capacity of the earth which is given as

C= 4×3.14 x(1/9×10⁹) x 6.4x 10⁶ = 711 x 10^-6F  or 711μF .

For next topic of electrostatics  which is parallel plate capacitor -click here –

Gauss’s law and proof of Gauss’s law .

In this topic we will learn Gauss’s law , derivation of gauss’s law , derivation of coulomb’s law using gauss’s law .

Before to know about Gauss’s law and proof of Gauss’s law student must know electrostatic flux , to read the notes on electrostatic flux click here-

Question in the examinations may be asked as –

state and prove Gauss’s law , using gauss’s law derive Coulomb’s law.

Gauss’s law in electrostatics –

According to Gauss for a closed surface total flux  in the free surface is 1/ϵ₀ times the total charge inside the closed surface.

To download the pdf of Gauss’s law and proof of Gauss’s law click here-

Electric flux and area vector

In this topic we will discuss about Electric flux and area vector .

question may be asked as what is electric flux ? write its unit and dimension.

Area vector and electric flux –

As we know area is a scaler quantity , but in some cases or in case of some problems area is taken as vector quantity . the direction of area is taken as the normal of the surface at a given point . The vector associated with every area element of a closed surface is taken to be in the direction of the outward drawn normal .

 

Electric flux – Electric flux over an area in an electric field is the total number of field line passes through that area normally .  It is denoted by ɸ its unit is Nm²C^-1 . And its dimension is  [M¹L³T^-3A^-1]. Electric flux is a scaler quantity.

The  number of electric field lines crossing this area is  proportional to field intensity E . Let electric field lines makes an angle ϴ with the area vector dS , and E is the electric field intensity , then we can write  dɸ =  E. (dS cosϴ ) [ since dScosϴ is the component of area vector normal to the surface ].

So ,  dɸ = E dS (With vector sign).

Or,  ɸ_E = ∫E dS cosϴ

To download the pdf of complete assignment related to Electric flux and area vector ( This pdf includes Gauss’s law and its applications). click here-