Exercise XI

 Maharashtra state Board 

Exercise Pdf notes 


1.  Units and Measurements  👈 Click on the link to download  pdf.

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Electrostatis

Electrostatics

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Move the balloon towards sweater and then towards wall and observe the phenomenon of electrostatics.









Electrostatics : A study of charges at rest is called as electrostatics.

In our day to day life we come across various object which is made up of atom . Atom consist of positively Charged Protons and Negatively Charged Electrons which revolve around the atom. This charges shows various characteristics when they are at rest as well as when they are in motion. In this chapter we are going to deal with charges at rest.

In earlier class we have seen various method of charging a body such as 
1) Conduction 

2) Induction  

Gauss Law

Gauss Law : The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ε0.  . It is a powerful tools which helps us to find total electric flux coming out of the closed surface  .

Mathematically it is given by
                                                        ∅ = q/ε0           ---------(1)
                                                            OR
                                                                   ∅ = ∫ E .ds ----------(2)
Where  ∅ is the total flux and q is the charge enclosed within that surface .

Application of gauss Law

1) Electric field intensity due to Uniformly charged  spherical shell or hollow sphere.


Consider a sphere of radius R  charged with a uniform charged density  σ in a dielectric medium ε. Consider a gaussian surface as a sphere with radius r surrounding the charged sphere.Consider a point P where electric field intensity is to be measured.

The total charge on a sphere is given by 
            q = σA
                                             q=σ 4πR---------------(3)
By gauss law, net electric flux is given by  ∅ = q/ε0  ------------from (1)    
Now due to symmetry , the electric field at each and every point is uniform  and directed radially outward and the direction of area is vector is parallel to direction of electric field .ie. θ= 0 .

We know that , Electric flux is given by,  
  ∅ = ∫ E .ds cos θ------------ from (2)
                            ∅ = ∫ E .ds cos0  ( θ=0)
           ∅ = ∫ E .ds  
                                ∅ =  E .4πr-------------(4)
                                                                   
          From (1) and (4)

                         q/ε0 = E .r2
                                      E =  q/ε0r2
                                     E =σ 4πR/ε0r2   ------------------from(3)
                                          Eσ R/ε0r2 -----------------(A)
                                      Equation (A)  gives Electric field intensity at point P 

                  Case 1: 
 if point P lies on the sphere then  R= r  , then equation (A) changes to 
                  Eσ  /ε0

                Case 2: 
if point P lies inside the sphere , then E= 0  ( electric field inside a sphere is always zero)
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Mathematical Physics

 Lesson No 02. Mathematical Physics

🕮 Introduction

In previous lesson we have studied about various physical quantities which can be measured.But all these quantities cannot be fully described by measurements and units only.There is a need to do perform mathematical operation to obtain a desired resultsThis physical quantities are described fully using magnitude,direction or both.In order to describe Physical Quantity , it has been categorized into two types:

a) Scalar Quantity

b) Vector Quantity

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