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Electrostatics-I
1. INTRODUCTION:
2. COULOMB’S LAW
Coulombs, through his experiment found out that the two charges ‘q_{1}’ and ‘q_{2}’ Kept at distance ‘r’ in a medium as shown in figure-1 exert a force ‘F’ on each other. The value of force F is given by
This law gives the net force experienced by q_{1} and q_{2} taking in account the medium surrounding them. Where F gives the magnitude of electrostatic force q_{1} and q_{2} are the magnitudes of the two interacting charges. K is electrostatic constant which depends upon the medium surrounding the two charges. This force F acts along the line joining the two charges and is repulsive if q_{1} and q_{2} are of same sign and is attractive if they are of opposite sign. Let us take some examples on application of coulomb's Law.
Electrostatics-II
Chapter Cover: Questions Based on electric flux for various configuration, Question on solid angle concept, Gauss law Based Questions.
If a surface is three dimensional we consider a small elemental area dS on this surface and direction of this elemental area vector is along the local normal of the surface at the point where elemental area is chosen as shown. Thus
Here â is the unit vector in the direction along the normal at elemental area dS
Flux = ϕ=∫E. , If E is constant, ϕ=E.A
In a uniform electric field shown in figure. If ϕ be the flux passing through an area S which is normal to the electric field lines, the value of electric field strength at this surfaces can be given as
Or flux through the surface can be given as
ϕ=ES
If in an electric field, surface is not normal as shown in figure. Here the are ABCD is inclined at an angle Ѳ from the normal to electric field. Here we resolve the area ABCD in two perpendicular components as shown in figure. One is S cos Ѳ, which is area ABC’D’ normal to electric field direction and other is S sin Ѳ, which is area CDC’D’ along the direction of electric field.
Here the total flux passing through the given area ABCD is same which is passing through its normal components S cos Ѳ, thus here the flux ϕ through the area can be given as
ϕ=ES cosѲ [S cos Ѳ=area of ABC’D’]
If we consider the direction of area vector normal of the area surface, as shown in figure, Ѳ would be the angle between S and E. Thus flux through the surface ABCD can be given as
ϕ=E.S
dϕ = Ed cos θ
The flux through the surface M can be given as
Electric flux through a circular Disc:
Figure shows a point charge q placed at a distance? From a disc of radius R. Here we wish to find the electric flux through the disc surface due to the point charge q. We know a point charge q originates electric flux in radically upward direction. The flux of q which is originated in cone shown in figure passes through the disc surface.
To calculate this flux, we consider an elemental ring on disc surface of radius X and width dx as shown.
Area of this ring (strip) is
dS=2πx dx
The electric field due to q at this elemental ring is given as
If dϕ is the flux passing through this elemental ring we have
dϕ= EdScosѲ
Total flux through the disc surface can be given by integrating this expression over the whole area of disc thus total flux can be given as
The above result can be obtained in a much simpler way by using the concept of solid angle and Gauss’s Law, shortly we’ll discuss it.
Electric Flux Through the Lateral Surface of a Cylinder due to a Point Charge:
Figure. Shows a cylindrical surface of length L and radius R. On its axis at its centre a point charge q is placed. Here we wish to find the flux coming out from the lateral surface of this cylinder due to the point charge q. For this we consider an elemental strip of width dx on the surface of cylinder as shown. The area of this strip is
dS = 2πR. Dx
The electric field due to the point charge on the strip can be given as
If dϕ is the electric flux through the strip, we can write
dϕ=EdScosѲ
Total flux through the lateral surface of cylinder can be given by integrating the above result for the complete lateral surface, which can be given as
The solution of above integration is left for students as exercise. This situation can also be easily handled by using the concepts of Gauss’s Law, we’ll discuss in next section.
Electric flux Produced by a Point Charge
The figure. shows a point charge placed at the centre of a spherical surface of radius R from which electric lines are originated and coming out of the surface of sphere. For clarity and convenience only lower half of sphere is drawn in the picture. As the charge q is inside the sphere, whatever flux it originates will come out from the spherical surface. To find the total flux, we consider an elemental area dS on surface. The electric field on the points on surface of sphere can be given as
The electric flux coming out from the surface dS is
Total flux coming out from the spherical surface is
At every point of spherical surface, magnitude of electric field remains same hence we have
Thus total flux, the charge q originates is . Similarly a charge –q absorbs
electric lines (flux) into it.
Figure shows a charge q enclosed in a closed surface S of random shape. Here we can say that the total electric flux emerging out from the surface S is the complete flux which charge q is originating, hence flux emerging from surface is
The above result is independent of the shape of surface if only depends on the amount of charge enclosed by the surface.
Flux Calculation in the Region of Varying Electric field:
In a region electric field depends on x direction as
E=E_{0}x^{2}
In the cube of edge a shown in figure from front face electric
Flux goes in which can be given as
ϕ_{in}=E_{0} (2a)^{ 2.}a^{2}=4E_{0}a^{4}
From the other surface flux coming out can be given as
ϕ_{out }= E_{0}(3a^{2}).a^{2}=9E_{0 }a^{4}
Here ϕ_{out}>ϕ_{in} for the cubical surface hence net
Flux=ϕ_{out}-ϕ_{in}=5 E_{0}a^{4}
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A large uniformly charged(negative) plate is placed in XZ plane and positve charge Q is fixed on the Y-axis. A dipole is placed in between these as shown. The dipole initially moves in
??
plz solve the question