ELECTRIC FIELD INTENSITY

ELECTRIC FIELD INTENSITY:-

A charged particle exerts a force on particles around it. We can call the influence of this force on surroundings as electric field. It can be also stated as electrical force per charge. Electric field is represented with E and Newton per coulomb (N/C) is the SI unit of it and its dimensions are [MLT^-3A^-1]

since electrical field is a vector quantity so in vector  form it can be written as

       Electric field is a vector quantity. And it decreases with the increasing distance

• · Electric field cannot be seen, but you can observe the effects of it on charged particles inside electric field.

• · To find the electric field vector of a charge at one point, we assume that as if there is a +1 unit of charge there.

• · If you want to find the total electric field of the charges more than one, you should find them one by one and add them using vector quantities

CONCLUSION:-

• Electric field intensity in inversely proportional to square of distance between two charge that is as the distance from the charge increases electric field intensity decreases

• Electric field intensity is inversely proportional to relative permittivity (e_r )  of  a dielectric i.e

  

NOTE:-

The point where electric field intensity is zero is called null point. It is in the straight line with the charges. In case of similar charges null point of neutral point lies between the charges

similarly in case of opposite charges if one charge is greater then the other then neutral point shits towards small charge instead of centre or we can say that neutral point lies behind smaller charge as shown in fig

 null point lies in the straight line joining the charges.

SOLVING PROBLEMS OF ELECTRIC FIELD:-

PROBLEM 1

A small object of mass 3.40 g and charge -15 µC "floats" in a uniform electric field. What is the magnitude and direction of the electric field?

SOLUTION:

If the object is floating, then the force exerted by the electric field exactly balances the force of gravity. The force of gravity is 

F = mg = (0.0034 kg)*(9.8 m/s^2) = 0.033 N 

The strength of an electric field is expressed in Newtons per Coulomb, meaning that the field exerts a force proportional to the charge of an object. If it balances gravity, 

E = F/Q = (0.033 N) / (-15 x 10^-6 C) = -2200 N/C

PROBLEM 2

Calculate vertically the net force on q.

SOLUTION:

First of all in order to calculate net force on q we will further direct forces in the fig

we give directions to the forces while just keeping in mind that force is directed from positive to negative. Since we know that angle formed by cutting two parallel lines are equal so the above angle is also θ1. By determining x and y coordinates we will get the fig as show above.There are two force acting on q one by q1 and other by q2 so net force on q is
F=F1+F2

We know net force on q can also be written as

F=Fx+Fy ----(i)

where Fx=F1x+F2x----(ii)
and Fy=F1y+F2y----(iii)
first of all we will calculate θ1 and θ2
for θ1 Tanθ1=0.8/0.6
θ1=53.13 degrees
similarly for θ2 Tanθ2=0.8/0.6
θ2=53.13 degrees

First of all we will calculate F₁ and F₂

For F₁ q₁=1x10^-6 and q₂= 4x10^-6so

 F₁=9x10⁹x1x10^-6x4x10^-6/1²

So F₁=0.036N

For F₂  q₁ = ­-1x10^-6 but magnitude of charge is 1x10^-6 and q₂=4x10^-6

So F₂=9x10⁹x1x10^-6x4x10^-6/1²

F₂=0.036N

F1x=F1cosθ=0.036xcos(53.13)=0.0216

F2x=F2 cosθ=0.036xcos(53.13)=0.0216

Similarly F1y=F1sinθ=0.036xsin(53.13)=0.0288

F2y=F2 sinθ=0.036xsin(53.13)=0.0288

putting value of F1x and F2x in (ii) we get 

Fx=0.0216+0.0216

Fx=0.0432

in fig we see that F1y and F2y are in opposite direction and tend to cancel each other so if F1y is positive then F2y is negative

so Fy=0.0288+(-0.0288)

Fy=0

putting value of Fx and Fy in (i) we get net force on charge q
so 

PROBLEM 3
Two positive point charge q1= -1 micro C and q2= 4 micro C are separated by a distance of 3 meter Find the spot on the line joining 2 charges where the Electric Field intensity is zero.
SOLUTION:

We know that in case of opposite charges neutral point exists behind the smaller charge hence the fig would be as shown in fig. Consider that neutral point lies at distance x m from the negative charge.and (3+x)m from positive charge. At point C we can say that electric field due to negative charge is equal to electric field due to positive charge equal to zero.
q1=-1 micro C but magnitude is 1 micro C and q2=4 micro C
we can write

                  E_A=E_B

after putting values we get 

1x10^-6/x²=4x10^-6/(3+x)²

By cross multiplying we get

10^-6.(3+x)²=4x10^-6.x²

After further cancellation we get

9+x²+6x=4x²

3x²-6x-9=0

After solving by using quadratic formulae of any other method we get

x=-1 and x=3

x can not be negative so x=3m

hence we can say that null point is 3m from negative charge or 6m from positive charge. 

PROBLEM 4:-

Three charges +q are placed at the three vertices of an equilateral triangle then the electric field at centroid of triangle would be.

SOLUTION:-

First of all one should know what is centroid. Basically centroid is the point of intersection of median and it divides the median in the ratio of 2:1 that is from fig in median AC if BC=x then AB=2x . Now if we move towards our question our we have on our three vertices three charges of equal magnitude and each side has magnitude a. Consider triangle ACD for an equilateral triangles medians are perpendicular so by applying pythagoras theorem we would easily calculate length of median AC would be a√3

                                                            2

since centroid divides the medians in the ratio of 2:1 so we conclude that AB=  a         and BC=   a                                                       

               √3                      2√3                                                                              fig (i)

 now consider fig (ii)  which is according to the question now what we are going to do is that  We will find the resultant of any two electric field intensity by adding the vertical and horizontal components here consider we that E₂ and E_{3. Distance of the vertices to ceteroid is same for all vertices which we can calculate easily consider right angled triangle BCD here hypotenuse=BD=distance of vertex to to centreoid=?}

_{perpendicular=BC=  a}     as proved above

                             2√3   

DC=base=a/2 (since every side is equal to a)

applying pythagoras theorem we find that BD=a/√3          

similarly EB also =a/√3                                                                                  fig (iii)

thus by using formulae E=kq/ r² 

so E₂ = E_{3 = 3kq/}r² 

medians of equilateral triangle also bisects the angle so for equilateral triangle every vertex has an angle of 60 degrees so angle BDC would be of 30 degrees also angle BEC=30 degrees

now we will find resultant of E₂ & E_{3  from fig (ii) we see that} E_2x & E_{3x cancel each other and} E_2y & E_{3y are add up so} E_2y = E_3y =(3kq/r²)sin30

so resultant of

E₂ & E₃₌E + E₃              

E_x+E_y

E_x=E_1x+E_2x=0

E_y=E_1y+E_2y=2(3kq/r²)sin30=3kq/r²

So resultant is equal to 3kq/r² and is in opposite direction to E₁ so resultant is 0