Electric Field Of Charged Spherical Shell
Gauss’s law can be used to measure the electric field of distributed charges like electric fields due to a uniformly charged spherical shell, cylinder , plate etc. In this article, we will use Gauss’s law to measure electric field of a uniformly charged spherical shell . Gauss’s law states that :
The net electric flux through any hypothetical closed surface is equal to (1/ε0) times the net electric charge within that closed surface
The hypothetical closed surface is often called the “Gaussian Surface”. According to Gauss’s law, if the net charge inside a Gaussian surface is Σq, then the net electric flux through the surface , φ = Σq/ε₀
Electric Field Of Charged Hollow Sphere
Let us assume a hollow sphere with radius r , made with a conductor. The conducting hollow sphere is positively charged with +q coulomb charges.
If the sphere has equal density all over its surface , then +q charge will be equally distributed all over the surface. So, the entire system is a symmetric system. From this symmetricity , we can say that the direction of the electric field will be radially outwards or inwards. Otherwise , the symmetricity will be lost. As the charges are positive , the sphere will repulse any positive point charge near it . So, the direction will be radially outwards. Also we can conclude that the magnitude of the electric field will be equal to the equidistant distances from the center because of the symmetricity.
Now, let us assume a hypothetical sphere with radius R and the same center as the charged sphere.
The outer spherical surface is our Gaussian Surface. The net charge inside the Gaussian surface , Σq = +q . According to Gauss’s Law, the total electric flux through the Gaussian surface ,
φ = q/ε₀ . . . . . (1)
Mathematically the flux is the surface integration of electric field through the Gaussian surface.
φ = ∮E . dA [dot product of E and dA]
or, φ = ∮E*dA*cos θ . . . . . (2)
From equation (1) and (2) ,
q/ε₀ = ∮E*dA*cos θ
E is constant through the surface . Because , all points on the surface are in same distance from the center. Previously in this article , we said that according to symmetricity, E will be constant in all equidistant places from the center. So, E can be brought out from the integration sign.
q/ε₀ = E ∮dA*cos θ …………..(3)
θ is always 0⁰ . As both the direction of dA and E are the same(radially outwards). So, the angle between them is 0.
So, equation (3) becomes ,
q/ε₀ = E ∮dA * cos 0⁰
q/ε₀ = E ∮dA * 1
q/ε₀ = E ∮dA
Now, ∮dA is the surface area of the outer sphere . So, ∮dA = 4πR²
q/ε₀ = E * 4πR²
So,
This expression is the same as that of a point charge. It is as if the entire charge is concentrated at the center of the sphere.
Electric Field On The Surface Of The Sphere (R = r)
On the surface of the conductor , where R = r , the electric field is :
E = (1/4πε₀) * (q/r²)
Electric Field Inside Hollow Sphere
If we assume any hypothetical sphere inside the charged sphere, there will be no net charge inside the Gaussian surface . So, Σq = 0 .
So, the net flux φ = 0.
So, ∮E*dA*cos θ = 0
Or, E ∮dA*cos θ = 0
Or, E = 0
So, the electric field inside a hollow sphere is zero.
Electric Field Of Charged Solid Sphere
If the sphere is not hollow , instead it is a solid one , then the entire charge will be distributed on the surface of the solid sphere. Because, in electrostatic condition , there is no electric field inside a conductor. (This topic is explained here : Electric Field Inside A Conductor). As there is no electric field inside a conductor , if we assume any hypothetical surface inside a conductor , the net flux φ will be zero. Thus, the net charge inside a conductor Σq = 0.
Thus , if +q charge is given to a solid sphere, it will be distributed equally throughout the surface of the sphere . There will be no charge inside the sphere. So the electric fields will be the same as the hollow sphere.
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