The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Question Id: PH0000006 

Question: The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Answer:

(a) Distance between the two spheres is 0.12m.

(b) Force on the second sphere due to the first sphere is save and i.e  0.2 N.

Explanation:

Given, 

two small charge \(q_{1}=0.4\mu C=0.4\times 10^{-6}C=4\times 10^{-7}C\) ,  \(q_{2}=-0.8\mu C=-0.8\times 10^{-6}C=-8\times 10^{-7}C\)  

and force(F)  between them is 0.2 N.

Distance between the sphere of charges is r (say)

force between 0.4 µC and –0.8 µC
Figure-1: force between 0.4 µC and –0.8 µC

According to Coulomb's law 

If two point charge \(q\) and \(Q\) are separated by a distance \(r\) then electrostatic force acting on them is \(F=\frac{1}{4\pi \epsilon }\frac{qQ}{r^{2}}\)     ;[where \(\epsilon\) is electric permittivity and  \(\epsilon =k\epsilon_{0}\) where k is dielectric constant and  \(\epsilon_{0}\) is electric permittivity for air medium]

Now Force between \(q\) and \(Q\) is 
\(F=\frac{1}{4\pi \epsilon _{0}}\frac{qQ}{r^{2}}\) ;[at air medium \(\epsilon =\epsilon_{0}\)  as \(k=1\)]
\[\Rightarrow 0.2=9\times 10^{9}\times \frac{(4\times 10^{-7}\times 8\times 10^{-7})}{r^{2}}\]
\[\Rightarrow r^{2}=9\times 10^{9}\times \frac{(4\times 10^{-7}\times 8\times 10^{-7})}{0.2}\]
\[\Rightarrow r^{2}=\frac{288\times 10^{9-14}}{0.2}\]
\[\Rightarrow r^{2}=\frac{288\times 10^{-5}}{2\times 10^{-1}}\]
\[\Rightarrow r^{2}=\frac{288}{2} \times 10^{-4}\]
\[\Rightarrow r^{2}=144 \times 10^{-2}\]
\[\Rightarrow r=\pm 12 \times 10^{-2}\]
\[\Rightarrow r=\pm 0.12\]
So, \(r=0.12m\) as distance is always positive. 
According to Coulomb's Law, electrostatic force of attraction or repulsion is same but opposite for each point charge. This is the Newtonian property of Coulomb's force. Hence, one can say that force on the second sphere due to the first sphere is save and i.e  0.2 N.

(a) Distance between the two spheres is 12m.
(b) Force on the second sphere due to the first sphere is save and i.e  0.2 N. (Answer)

What is the force between two small charged spheres having charges of 2×10−7C and 3×10−7C placed 30cm apart in air ?

Question Id: PH0000005

Question: What is the force between two small charged spheres having charges of \(2\times 10^{-7}C\) and \(3\times 10^{-7}C\) placed 30 cm apart in air?
Answer: 
Force acting  between \(q\) and \(Q\) is \(6\times 10^{-4}N\)

Explanation: 
Given two charge \(q=2\times 10^{-7}C\) and  \(Q=3\times 10^{-7}C\)
And they are separated by a distance \(r=30cm \Rightarrow r=0.3m\)

Figure-1


According to Coulomb's law 
If two point charge \(q\) and \(Q\) are separated by a distance \(r\) then electrostatic force acting on them is \(F=\frac{1}{4\pi \epsilon }\frac{qQ}{r^{2}}\)     ;[where \(\epsilon\) is electric permittivity and  \(\epsilon =k\epsilon_{0}\) where k is dielectric constant and  \(\epsilon_{0}\) is electric permittivity for air medium]
Now Force between \(q\) and \(Q\) is 
\(F=\frac{1}{4\pi \epsilon _{0}}\frac{qQ}{r^{2}}\) ;[at air medium \(\epsilon =\epsilon_{0}\)  as \(k=1\)]
\[\Rightarrow F=9\times 10^{9}\times \frac{2\times 10^{-7}\times 3\times 10^{-7}}{0.3^{2}}\]
\[\Rightarrow F=54\times \frac{10^{-6}}{0.09}\]
\[\Rightarrow F=6\times 10^{-4}\]

Hence, force acting  between \(q\) and \(Q\) is \(6\times 10^{-4}N\). (Answer)





Theory related to this problem:

Condition for applying Coulomb's Law:
(i) Only applicable for point charges that means size of the point charge is negligible with respect to their distance.
(ii) Applicable for the distance larger than nuclear range ( \(\approx 10^{-15}m\) )

Coulomb's law (Theory): Attraction force or repulsion force between between two point charge is proportional to the product of the charge and inversely proportional to the square of the distance between the charge.
i.e \(F\alpha (q_{1}\times q_{2})\)  and \(F\alpha \frac{1}{r^{2}}\)
So, \(F\alpha \frac{(q_{1}\times q_{2})}{r^{2}}\)
\(\Rightarrow F=k' \frac{(q_{1}\times q_{2})}{r^{2}}\)  ; [where k' is proportionality constant]

In C.G.S Unit: \(k'=\frac{1}{k}\)  [where k is electric permittivity]
Value of k=1 for air ( vaccum ) medium.
So, for C.G.S medium \(F=\frac{(q_{1}\times q_{2})}{r^{2}}\)

In S.I Unit: \(k'=\frac{1}{4\pi \epsilon }\)  [ \(\epsilon\) permittivity of medium]
And \(\epsilon =k\epsilon _{0}\)  where \(\epsilon _{0}\) is permittivity of air medium and k is dielectric constant.
So, \(F=\frac{1}{4\pi \epsilon _{0}}\times \frac{(q_{1}\times q_{2})}{r^{2}}\)

Value of  \(\epsilon _{0}=8.854\times 10^{-12} C^{2}N^{-1}m^{-2}\)
And value of  \(\frac{1}{4\pi \epsilon _{0}}=9\times 10^{9} Nm^{2}C^{-2}\) 

Hence in S.I \(F=9\times 10^{9} \times \frac{(q_{1}\times q_{2})}{r^{2}}\) in N unit.

Express \(Sine(sin)\), \(Cosine(cos)\) and \(tangent(tan)\) in terms of sides of triangle

Question: Express \(Sine(sin)\), \(Cosine(cos)\) and \(tangent(tan)\) in terms of sides of triangle 

Answer:
For a right angle triangle  we can find sine or sin, cosine or cos, tangent or tan in terms of their side. 
To know more accurately about sin, cos and tan we must have to know about (1)base, (2)height (perpendicular) and (3)hypotenuse.

Figure-1 (right angle triangle)

   In general we use base, hypotenuse and height(perpendicular) as shown in figure.  But this method is not always right.   Let's know what is right

(1) Definition of hypotenuse:     
The side opposite to the right angle or \(90^{o}\) is called hypotenuse, it is the largest side of right angle triangle. 
(2) Definition of perpendicular: 
It depends on the angle which we are going to consider. If we consider angle C the perpendicular is AB. 
If we consider angle A then perpendicular is  BC. (See below figure-2) 
so height is the opposite side of the angle which we have consider. 
(3) Definition of base:
Then the rest side is called base. 

Figure-2


Now,   
One can say that [considering angle \(\angle C\) ]
(1) Sine is the ratio of perpendicular and hypotenuse
\[sin C=\frac{perpendicular(AB)}{hypotenuse(AC)}\]

(2) cos is the ratio of base and hypotenuse
\[cos C=\frac{base(BC)}{hypotenuse(AC)}\]

(3) tan is the ratio of base and perpendicular
\[tan C=\frac{perpendicular(AB)}{base(BC)}\]

Again,
One can say that [considering angle \(\angle A\) ]
(1) Sine is the ratio of perpendicular and hypotenuse
\[sin C=\frac{perpendicular(BC)}{hypotenuse(AC)}\]

(2) cos is the ratio of base and hypotenuse
\[cos C=\frac{base(AB)}{hypotenuse(AC)}\]

(3) tan is the ratio of base and perpendicular
\[tan C=\frac{perpendicular(BC)}{base(AB)}\]

So, in this way one can easily get \(sin, cos, tan\) ratio. And using this one can find \(cosec, sec, cot\). 
[As \(cosec\theta =\frac{1}{sin\theta }\) ; \(sec\theta =\frac{1}{cos\theta }\) and \(cot\theta =\frac{1}{tan\theta }\) ] 

A right angle triangle has a base of \(25m\) and its area is \(375m^{2}\) find its height

Question: A triangle has a base of \(25m\) and its area is \(375m^{2}\) find its height? 

Answer:
Height of the right angle triangle is \(30m\) .
Explanation:
Given base is \(25m\) and area is \(375m^{2}\)

Figure-1 (right angle triangle)



Now we know area of right angle triangle= \(\frac{1}{2}\times base \times height\)
So according to question, 
\(\frac{1}{2}\times base \times height\)=375
\[\Rightarrow base \times height= 375 \times 2\]
\[\Rightarrow 25 \times height= 750\]
\[\Rightarrow height= \frac{750}{25}\]
\[\Rightarrow height= 30\]

Hence, height of the right angle triangle is \(30m\) . (Answer)

Project on atomic model and history || Class 12 physics Project on atomic model and history

Project on atomic model and history:

           Atom is the smallest constituent unit of ordinary matter that constituent of chemical elements viz electron, proton, neutron. Atoms are extremely small, typical sizes are around 100 picometers. In 1800 AD Jhon Dalton use the concept of atom to explain why chemical elements seemed to combine in ratios of small whole numbers. And Dalton give the atomic theory famous as "Dalton's atomic theory". Atom is mainly consist of two main part as nucleus and electron shell. The nucleus is made of proton and neutron. Proton is positively charged and neutron has no charge i.e neutral. And electron is negatively charge.

diagram of electron and proton
diagram of electron and proton

Discovery of Electron:
     In 1897, J J Thomson  discovered that cathode rays are not electromagnetic waves but are made of particles taht are 1800 times lighter than hydrogen (the lightest atom). And that particle are electron and it is negatively charged. The symbol used for electron is  or  . The electron has charge value  C. And mass is  kg . 

Discovery of  Proton: 
      In 1886, Eugen Goldstein observed proton as . In 1917 Ernest Rutherford identified and named Proton. Proton is positively charged paricle having charge opposit of electron i.e   C. Mass of the proton is  kg. Symbol used for proton p or 

Discovery of Neutron:
        In 1920, Ernest Rutherford theorized about neutron but after 12 year on 1932 his student James Chadwick discovered neutron. It is a neutral particle  i.e have no charge . Mass of neutron is  kg. Most heavy particle in atom. Its is responsible for most of the mass of an atom.

  Atomic model was first introduced by Jhon Dalton his statements were proved wrong. It does not matter that his model was not correct but he gave the first concept of atomic model and from his concept later more scientist developed atomic model.

    Before Neils Bohr , Rutherford gave an atomic model but is was also need correction . According to Rutherford's nuclear model of the atom (it is a classical concept) : "atom is an electrically neutral sphere consisting of very small, massive, and positively charged nucleus at the cent surrounded by the revolving electron in their dynamically orbit".
But  according to electromagnetic theory moving charge particle radiate electro magnetic radiation and that is nothing but and energy. So at some certain time the moving electron will loss his all energy and it will fall on the nucleus in a spiral way. Hence there will arise a question of existence of the atom. That's why this theory was declined by the scientist. Then Bhor Model gave the model of a stable atomic model.  

Bohr's atomic model: 
    It is some time also called as Rutherford-Bohr model. In 1913 the model was prepared by Neils Bohr and Rutherford postulate of this model are as follows

     1) The electron is able to revolve in certain stable orbits around the nucleus without radiating any energy, contrary to what classical electromagnetism  suggests. These stable orbits are called stationary orbits and are attained at certain discrete distances from the nucleus. The electron cannot have any other orbit in between the discrete ones.
                                     
    2)The stationary orbits are attained at distances for which the angular momentum of the revolving electron is an integral multiple of the  Plank's constant: , where n = 1, 2, 3, ... is called the principle quantum number, and ħ = h/2π. The lowest value of n is 1; this gives a smallest possible orbital radius of 0.0529 nm known as the Bohr radius. Once an electron is in this lowest orbit, it can get no closer to the proton. Starting from the angular momentum quantum rule, Bohr was able to calculate the energies of the allowed orbit of the hydrogen atom and other hydrogen like atoms and ions. These orbits are associated with definite energies and are also called energy shells or energy levels. In these orbits, the electron's acceleration does not result in radiation and energy loss. The Bohr model of an atom was based upon Planck's quantum theory of radiation.

   3)Electrons can only gain and lose energy by jumping from one allowed orbit to another, absorbing or emitting electromagnetic radiation with a frequency ν determined by the energy difference of the levels according to the Plank relation: , where h is Plank's constant.







Project Report on semiconductors || Class 12 physics project on semiconductor

Project Report on semiconductors:-

INTRODUCTION:
        Most of the solids can be define into three classes: 1) conductor 2) insulator and  3) semiconductor (1) conductor: those solids can be called as conductor who have good electric conductivity property. Conductivity define by the presence of free electron. That means conductor have free electron by which it can easily conduct current. Example:- copper, iron etc. (2) insulator   those solid can be called insulator who have no electric conductivity. That means insulator have no free electron and hence they cannot conduct current. Example:- wood, glass etc. (3) semiconductor semiconductor are those solid whose electric conductivity is between conductor and insulator. Semiconductor can conduct current in some particular conditions. 

 Theory:-
           The energy band structure of the semiconductors is similar to the insulators but in their case, the size of the forbidden energy gap is much smaller than that of the insulator. In this class of solid  the forbidden energy gap is of the order of about 1ev, and the two energy bands are distinctly separate with no overlapping. At absolute zero temperature, no electron has any energy even to jump the forbidden gap and reach the conduction band. Therefore the substance is an insulator. But when we heat the solid,  this provide some energy to the atoms and their electrons, it becomes an easy matter for some electrons to jump the small (» 1 ev) energy gap and go to conduction band. Thus at higher temperatures, the solid becomes a conductors. This is the specific property of the solid  is known as a semiconductor.



Effect of temperature on conductivity of Semiconductor:-
           At 0K, all semiconductors are insulators. The valence band at absolute zero is completely filled and there are no free electrons in conduction band. At room temperature the electrons jump to the conduction band due to the thermal energy. When the temperature increases, a large number of electrons cross over the forbidden gap and jump from valence to conduction band. Hence conductivity of semiconductor increases with temperature.
  • semiconductor can be divided into two types such as 1) intrinsic and 2) extrinsic
 
INTRINSIC SEMICONDUCTORS:
         Pure semiconductors are called intrinsic semi-conductors. In a pure semiconductor, each atom behaves as if there are 8 electrons in its valence shell and therefore the entire material behaves as an insulator at low temperatures.
       A semiconductor atom needs energy of the order of 1.1ev or more to shake off the valence electron. This energy becomes available to it even at room temperature. Due to thermal agitation of crystal structure, electrons from a few covalent bonds come out. The bond from which electron is freed, a vacancy is created there. The vacancy in the covalent bond is called a hole i.e hole is absence of electron.
     This hole can be filled by some other electron in a covalent bond. As an electron from covalent bond moves to fill the hole, the hole is created in the covalent bond from which the electron has moved. Since the direction of movement of the hole is opposite to that of the negative electron, a hole behaves as a positive charge carrier. Thus, at room temperature, a pure semiconductor will have electrons and holes wandering in random directions. These electrons and holes are called intrinsic carriers.
      As the crystal is neutral, the number of free electrons will be equal to the number of holes. In an intrinsic semiconductor, if   denotes the electron number density in conduction band,  the hole number density in valence band and  the number density or concentration of charge carriers, then

                                         

EXTRINSIC SEMICONDUCTOR:
       As the conductivity of intrinsic semi-conductors is poor type, so intrinsic semi-conductors are of little   practical importance. The conductivity of pure semi-conductor can, however be enormously increased by addition of some pentavalent or a trivalent impurity in a very small amount (about 1 to 106 parts of the semi-conductor). The process of adding an impurity to a pure semiconductor so as to improve its conductivity is called doping. Such semi-conductors are called extrinsic semi-conductors. Extrinsic semiconductors are of two types :

            i) n-type semiconductor

            ii) p-type semiconductor

   i) n-type semiconductor

                      
                                 (black dot indicate "electron" and white dot indicate "hole")
          When an impurity atom belonging to group V of the periodic table like Arsenic is added to the pure semi-conductor, then four of the five impurity electrons form covalent bonds by sharing one electron with each of the four nearest silicon atoms, and fifth electron from each impurity atom is almost free to conduct electricity. As the pentavalent impurity increases the number of free electrons, it is called donor impurity. The electrons so set free in the silicon crystal are called extrinsic carriers and the n-type Si-crystal is called n-type extrinsic semiconductor. Therefore n-type Si-crystal will have a large number of free electrons (majority carriers) and have a small number of holes (minority carriers).
        In terms of valence and conduction band one can think that all such electrons create a donor energy level just below the conduction band as shown in figure. As the energy gap between donor energy level and the conduction band is very small, the electrons can easily raise themselves to conduction band even at room temperature. Hence, the conductivity of n-type extrinsic semiconductor is markedly increased.
        In a doped or extrinsic semiconductor, the number density offree electron in the conduction band () and the number density of holes in the valence band () differ from that in a pure semiconductor. If   is the number density of electrons is conduction band, then it is proved that

                                                =

ii) p-type semiconductor

                    
                             (black dot indicate "electron" and white dot indicate "hole")
      If a trivalent impurity like indium is added in pure semi-conductor, the impurity atom can provide only three valence electrons for covalent bond formation. Thus a gap is left in one of the covalent bonds. The gap acts as a hole that tends to accept electrons. As the trivalent impurity atoms accept electrons from the silicon crystal, it is called acceptor impurity. The holes so created are extrinsic carriers and the p-type Si-crystal so obtained is called p-type extrinsic semiconductor. Again, as the pure Si-crystal also possesses a few electrons and holes, therefore, the p-type si-crystal will have a large number of holes (majority carriers) and a small number of electrons (minority carriers).
     It terms of valence and conduction band one can think that all such holes create an accepter energy level just above the top of the valance band as shown in figure. The electrons from valence band can raise themselves to the accepter energy level by absorbing thermal energy at room temperature and in turn create holes in the valence band.
    Number density of valence band holes () in p-type semiconductor is approximately equal to that of the acceptor atoms () and is very large as compared to the number density of conduction band electrons (). Thus,

                                    >>  >> 

ELECTRICAL RESSITIVITY OF SEMICONDUCTOR:
      Consider a block of semiconductor of length L, area of cross-section A and having number density of electrons and holes as  and  respectively. Suppose that on applying a potential difference, say V, a current I flows through it as shown in figure. The electron current () and the hole current () constitute the current I flowing through the semi conductor i.e.

                                      +      ........ (i)

                        If  is the number density of conduction band electrons in the semiconductor and , the drift velocity of electrons then  

                        Similarly, the hole current, 

                        From equation    (i) I=

                                                            I=       ........(ii)

                        If r is the resistivity of the material of the semiconductor, then the resistance offered by the semiconductor to the flow of current is given by :

                                    R = r I/A         ........(iii)

                        Since V = RI, from equation (ii) and (iii) we have

                        

                            .......... (iv)

If E is the electric field set up across the semiconductor of unit cross section, then:

                              .......... (v)

            from equation (iv) and (v), we have

                                    

                                    

            On applying electric field, the drift velocity acquired by the electrons (or holes) per unit strength of electric field is called mobility of electrons (or holes). Therefore,

  Mobility of electrons and holes is given by :
      Mobility of electron is   and mobility of hole is 

                          and  

                             ......... (vi)

Also,     is called conductivity of the material of semiconductor

                            ....... (vii)

                        The relation (vi) and (vii) show that the conductivity and resistivity of a semiconductor depend upon the electron and hole number densities and their mobilities. As  and  increases with rise in temperature, therefore, conductivity of semiconductor increases with rise in temperature and resistivity decreases with rise in temperature.

                           Different Types of Semiconductor Devices
           Now in modern age most of the electronics device is composed by semiconductor device as memory card, chips, pendrive, computer, mobile and so many digital instrument we use in our everyday life.

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