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PHYSICS-11

Chapter 29 - Electrostatics

Definations:

1. Electric field strength: It is defined as the field or space upto which force due to electric charges is exert. Mathematically, electric field strength is electrostatic force per unit charge.
= /q
= (1/4πε0)(q.q1/r2.q)
therefore, E= (1/4πε0)(q1/r2)

if test charge is very small then,
E= q 0 (F/q)

2. Superposition of electric field intensity: If there are various electric field intensity due to multiple charges then total intensity can be calculated by the vector sum of various intensity.
Let, E1, E2, E3, ..... are various electric field acting at a source charge then total intensity is
= 1 + 2 + 3 + 4 + .....

Electric flux

It is defined as electric lines of force/electric field passing through a surface area held perpendicular to the direciton of electric field.
Let, 'E' be an electric field due to any charge at a point then if it passes through a surface area A which is perpendicular to the field then electric flux (ɸ) is defined as,
ɸ= E.A <------ (i)

Electric flux due to a point charge:
Consider a point charge Q with electric field E which exerts electric force uniformly in all direction. Then from definition of electric flux,
(1/4πε0)(q1/r2)= E.A
ɸ= E x 4πr2 <------ (ii)
Also we have,
Electric field intensity, E= (1/4πε0)(q/r2) <------- (ii)
From equation (ii) and equation (iii),
ɸ= (1.q/4πε0r2)
ɸ= q/ε0
which is the electric flux due to the point charge in the space.

Gauss law

The total flux passing through any closed surface is equal to 1/ε times of the total charge enclosed in the surface.
Mathematically, it is the ratio of total charge present in the given surface to permittivity.
ɸ= Q/ε (for medium)
ɸ= Q/ε0 (for air/space)

Application of Gauss law:
1. Electric field due to point charge at surface of sphere.



Consider a sphere having radius 'r' and 'q' be the charge present at any point of the sphere (+ve charge) here we are going to find the electric field intensity at the surface of this sphere due to charge 'Q'.
From the defination of electric flux,
ɸ= E.A
ɸ= E.4πR2 <-------- (i) [where Area=4πr2]
Again, from the definition of gauss law,
ɸ= Q/ε0 <------- (ii)
from equation (i) and equation (ii)
Q/ε0 = E.4πR2
E= (Q/ε04πR2)

therefore, E= (1/4πε0).(Q/R2) <------- (iii)

2. Electric field due to point charge outside the sphere:



Consider a sphere of radius R. Let, r be the distance from center of sphere to the point outside the sphere where electric field is to be calculated.
From the definition of electric flux,
ɸ= E.A [where E is electric field]
A= area of sphere upto the distance 'r' outside the surface of sphere
ɸ= E.4πr2 <------ (i)
Also, from gauss law,
ɸ= Q/ε0 <------- (ii)
from equation (i) and equation (ii)
Q/ε0 = E.4πr2
therefore, E= (Q/4πr2ε0)

Conclusion: This equation indicates that the charge is uniformly distributed over the surface of sphere. It behaves as if the whole charge is concentrated as the centre of sphere.

3. Electric field inside the sphere:



Consider a sphere of radius 'r'. Let, Q be the uniformly distributed charge over the sphere. Now we have to calculate electric field intensity inside the sphere.

From the definition of electric flux,
ɸ= E.A [where E is electric field]
A= area of sphere and ɸ is electric flux.
ɸ= E.4πR2 <------ (i)
Also, from gauss law,
ɸ= Q/ε0 <------- (ii)
from equation (i) and equation (ii)
Q/ε0 = E.4πR2
E=0 then, Q=0

No charges are present inside the sphere. Therefore total electric field inside the sphere is always 0.

4. Electric field outside plane charged sheet:



Consider a plane charged sheet, where a circle is drawn outside where the electric field is to be calculated. The electric field is uniform in all part of this circle. Since, electric field is perpendicualr to the charged sheet but parallel to the region where electric field intensity is to be calculated.
Let, ɸ be an electric flux due to charges then,
ɸ= E.A <------ (i)
where A is area of cylinder
Again from gauss law of electrostatics,
ɸ= Q/ε0 <------- (ii) [where Q is charge present and ε0 is permittivity in space.]
From equation (i) and equation (ii)
E.A= Q/ε0
E= Q/ε0A <------(iii)

From the definition of surface charge density σ= Q/A
A=Q/σ
therefore, E=σ/ε0


5. Electric field due to one dimensional (linear) charged rod:



Consider an infinite charged conducting rod of linear charge density (λ). Let, 'p' be any point 'r' distance away from conducting rod. By symmetry at any all other point at same distance from conducting rod, electric field intensity is uniform. So, we have considered a cylinder of length 'L' as shown in figure above.
Now applying Gauss law,
ɸ= Q/ε0 <--------- (i)
Also from definition of electric flux
ɸ= E.A
From equation (i) and equation (ii)
E.A= Q/ε0
therefore, E= λ/2πε0r

6. Electric field intensity due to infinite plane sheet of charge:



Consider a plane sheet of surface charge density 'σ'. Let, 'P' be any point at distance 'r' from sheet where electric field intensity is to be calculated. A cylinder of area 'A' is constructed as shown in figure. By symmetry magnitude of electric field at distance 'r' in opposite side of 'p' is same as at point P.
Applying Gauss theorem,
electric flux (ɸ)= Q/ε0 <---------(i)
Again from definition of electric flux,
ɸ= E2A <-------- (ii) [2A for considering both side of charged sheet]
From equation (i) and equation (ii)
E2A= Q/ε0
[Q= σ.A] E= Q/2Aε0
[σ= Q/A] E= σ/2ε0

7. Force experienced by charged particle in electric field:



A negatively charged oil drop is placed in between two parallel plates A and B as shown in figure above. Electric potential 'V' is supplied is placed in plates A and B so force is experienced by charged oil drop in between the parallel plates.
Let 'Q' be the charged of supplied electric field then force experienced by charged oil drop due to electric field,
F= Q.E <--------- (i)
Since, the weight of charged oil drop is acting vertically downward.
therefore, downward acting force (W)= m.g <-------- (ii)
For the charged drop remain to be stationery in between the parallel plates upward acting force must balance downward acting force.
therefore, QE= mg
Q.(V/d)= volume(v) x ρ x g [E=V/d] and [m= v x ρ]
Q.(V/d)= 4/3(πr3.ρ.g)
Q= 4/3V(πr3.ρ.d.g) <--------- (iii)

if 'n' be the number of electron in given potential then,
Q= ne
ne= 4/3V(πr3.ρ.d.g)
therefore, n= 4/3eV(πr3.ρ.d.g)

COULUMB'S LAW
1. Columb's law of force of attraction or force of repulsion between two charges: The force of attraction or repulsion between two charges is defined by columb's law considered q1 and q2 be two charges 'r' distance apart from each other.
F ∝ (q1.q2)/r2
F= K(q1.q2)/r2 [where 'k' is proportionality constant]
K= (1/4πε0) [where 'ε0' is permittivity of free space]

Therefore, F= (1/4πε0) (q1.q2)/r2

2. Permittivity: The permittivity explains the respnose of given medium to the charges present on it. The unit of permittivity is
ε0= N-1m-2C2
ε0= C2/Nm2
ε0= C2N-1m-2
And it's value in air is ε0= 8.85 x 10-2 C2N-1m-2

3. Relative permittivity (εr): Relative permittivity is the ratio of permittivity of any medium to permittivity in its space and it is called as dielectric constant.
Mathematically, Dielectric constant is a ratio of the columb's force between two charges in air to columb force between the same charges in any other medium.
Dielectric constant (k)= (1/4πε0)[q1.q2)/r2]
= {(1/4πε0)[q1.q2 r>)/r2]} / (1/4πε0)[q1.q2)/r2]
therefore, K= εr

Electrostatic Induction:

THe temporary electrification of conduction by bringing a charged body by closed to it is called Electrostatic Induction.
For eg: charging a body by friction, by conduction, or by induction.

(a.) By friction:
In this method, two bodies are rubbed against each other due to which there is transfer of charge(electrons) between these two bodies so that, one becomes positively charged and another becomes negatively charged. In this case, nature of the charge depends upon the materials used.

(b.) By conduction:



The process of charging a body by touching it with a charged body is called charging by conduction. In this method, the charge produce on the conductor is of the same nature as is on the charging body as shown in figure:

Conductor A is positively charged. So, it has a deficiency of electron. When it is brought in contact with a neutral conductor B, the free electrons of B will flow to A to makeup for its deficiency. Due to loss of electron in B, it becomes positively charged. On the other hand, if B is to be negatively charged, a negatively charged conductor A is made to touch it. In this case transfer of electron takes place from A to B.

(c.) By induction:



Charging a body positively by induction:
An uncharged body can be charged positively by following ways:
i. Considered as uncharged body (sphere) having two sides A and B as shown in figure:

Step1:



A charge rubber rod kept near to side A of the sphere, all positively charges are attracted towards the rod as shown in figure above. Here side A is filled with positive charges and side B is filled with negative charges.

Step2:



With the help of a conducting wire side B is grounded. So, all the negative charges move towards earth. Hence, only positive charges remaining in the sphere are being attracted towards the rubber rod which is shown in figure above.

Step3:



Now the conducting wire is remove from the sphere, only positive charges remained on a sphere attracting towards the rubber rod as shown in figure above.

Step4:



Finally, rubber rod is removed from side A of the sphere. Now, positive charges spread all over the sphere as shown in figure above.

Linear charge distribution

It is defined as total charge present on distributed in 1 dimensional body per unit length. Therefore, linear charge distribution is equal to:
Total charge present/length
Q/L columb per metre

Surface charge distribution

It is defined as total charge present in two dimensional body per unit area of body. Therefore surface charge distribution is equals to:
(total charge present) / area
Q/A C/m2 (columb per meter square)

Volume charge distribution

It is defined as total charge present in three dimensional body per unit volume of a body. Therefore volume charge distribution is equals to :
total charge present/volume
Q/V C/m3 (Columb per meter cube)
The distribution of charge depends on instructure and shape of a body. It is also affected by a conduction present near to it.
The greater the curvature at any point the greater will be the accumulation of the charge at that point. In case or rectangular shape of body, the charge density is more at corners. Similarly, in case of conical charge body maximum charges are present at apex.,


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