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The presence of charged particles anywhere in space creates an electric field. Wherever you find a charge, the electric field must be there. Thus, An electrical property associated with each point in space when charge is present in any form.. In the presence of another charge, they will exert force on each other, and this force is the electric force. The strength of the electric field E, at any point, can be defined as force per unit positive charge, i.e., E = F/Q. This is known as electric field intensity, and its SI unit is volts/metre. Electric field intensity depends on the source charge and not on other charges present in the vicinity. For example, if we bring a test charge having twice the magnitude of charge, the electric force will also be double, but the ratio of F/Q will remain constant. That means E will remain stable at that point. The involvement of the test charge slightly changes the existing field.
Electric Field Intensity
Electric field intensity is a vector quantity that has both magnitudes and direction. The direction of the force depends on the nature of the charge. For example, the direction of a negative charge would be in the opposite direction of the electric field. The direction of positive charge is conventionally taken to be the direction of the electric field. The field produced by an isolated positive charge is outward as it always repels each other. Electric field lines represent an electric field, and they always originate from the positive charge (source) and end at the negative charge (sink). The electric field decreases as the distance increases because it’s inversely proportional to the square of the distance and direction to the magnitude of charge.
Electric Flux
As already discussed, the electric field is represented by electric field lines. The number of these lines that pass through a given area is called electric flux. Flux can be positive, negative, or zero. If the number of lines entering a surface is equivalent to the number of lines leaving it, then the total flux is zero. Electric flux is represented by and is given by
=E.A=EAcos
There exists a relationship between electric flux, charge, and permittivity of free space governed by a law known as Gauss’s law. Gauss’s law states that the total electric flux over any closed surface is equal to the ratio of charge enclosed and permittivity.
total=Q/0
The integral form of Gauss’s law is as follows
E.da=Q/0
Electric Field Intensity due to a Uniformly Charged Sphere
Let’s consider a uniformly charged non-conducting sphere of radius R and charge Q distributed equally inside it. Three cases arise:
Electric field intensity on the surface of this sphere, electric field inside it, and electric field intensity outside the sphere. The direction of electric field intensity on the Gaussian surface is radially outward which is in the direction of the area vector of the Gaussian surface.
Let’s take the first case where we need to calculate electric field intensity outside the sphere. Let’s consider an arbitrary point r outside the sphere. We need to find the electric flux using Gauss’s law. For that, draw a Gaussian surface at a point A outside the sphere such that its area element is dA. The direction of the area element is always perpendicular to a given surface and the direction of the electric field will be radially outward in this case. Therefore, =0.
Now, using definition of the flux
dE=E.dA=EdAcos=EdAcos0=EdA … 1
Integrating equation 1, we have
E=EdA … 2
Using Gauss’s law
E=Q/0 … 3
From 2 and 3
EdA=Q/0
Or, EdA=Q/0
Area of our chosen Gaussian surface would be the surface area of a sphere, which is 4r2.
E4r2=Q/0
E=Q/4r20 … 4
The charge will be distributed over the volume of the sphere. Hence, we have to calculate volume charge density.
Q=4/3*R3 … 5
Putting the value of Q in equation 4
E=(1/40)4R3/3r2
Therefore, E=(/0)R3/3r2
The above equation represents the value of the electric field outside the uniformly charged non-conducting sphere.
Electric field intensity on the surface of the uniformly charged non-conducting sphere.
We had earlier considered point A, outside the sphere. Now, imagine the same point A is located on the surface of the sphere such that R = r.
Replacing r by R in equation number 4, we will obtain
E=Q/4R20
Substituting the value of Q from equation 5 in the above equation
E=R/30
Electric field intensity inside the non-conducting sphere.
Now imagine the same point A is inside the sphere at a distance of r from the centre.
Our Gaussian surface has become smaller in size, which implies that the number of lines passing through it also decreases.
Therefore, E=E.4r2
Where E=q/0
Equating the two equations
E.4r2=q/0
Note that in the first two cases the enclosed charge was Q but now it has changed to q.
Volume charge density =Q/(4R3/3)=q/(4r3/3)
q=Q(r/R)3
Substituting the value of q in equation 4
E=Qr/40R3
Where q=Q(r/R)3
E=r/30
Conclusion
To summarise the concepts, the strength of the electric field E, at any point, can be defined as force per unit positive charge, i.e, E = F/Q. This is known as electric field intensity. The electric field intensity of a uniformly charged sphere is maximum on the surface of the sphere and zero at the centre of the sphere. As we move outside the sphere, intensity decreases according to its inverse square relation with distance, i.e., E=1/r2. Hence the sphere is electrically conductive.
