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We can define a spherical shell as a region between two concentric spheres having different radii. We can use Gauss’s law to calculate the electric field of a spherical shell. Gauss’s law is helpful in describing the relationship between electric charge distribution and the resulting electric field. The law states that the total number of electric flux that travels through any closed surface will be proportional to the electric charge contained in it. We use a Gaussian surface in order to simplify the calculations. The electric field of the spherical shell can be measured in two ways, one is Electric Field Outside the Spherical Shell and the other is Electric Field Inside the Spherical Shell.
The Electric field is caused because of the uniformly charged spherical shell.
The formula used to calculate the electric field is E = F/q. The force exerted on positive and negative charges are in opposite directions. Electric charges and magnetic fields that vary with time generate an electric field. The electric field has magnitude and direction, and hence, it is a two-dimensional vector quantity. Field lines are useful in depicting the electric field, and they are also known as force lines.
The electric field that is produced by the spherical shell can be measured in two ways i.e. Electric Field Inside the Spherical Shell and Electric Field Outside the Spherical Shell.
Gauss law states that the total electric flux that a closed surface has is equivalent to 1/ε0 times the total charge that a closed surface contains. The total charge that a closed surface contains is proportional to the total electric flux enclosed inside the surface. Hence, if we assume φ to be the total flux and ε0 is the electric constant, then the total charge enclosed by the surface would be:
Q = φ ε0
Therefore, when we express the formula of Gauss law in terms of charge, we write it as,
φ = QεO
where Q denotes the charge enclosed inside the surface and ε0 denotes the electric constant. Gauss law is an important law that is helpful in determining the electrical field of an object. It has many applications in the field of electromagnetism.
- Electric Field Outside the Spherical Shell
If we imagine a point that is outside the spherical shell and the distance between the point p and the centre of the spherical shell is r. We will use a Gaussian spherical surface for symmetry whose radius is r and its centre is O. As every point is equally distant “r” from the centre of the sphere, it can be seen that the Gaussian surface would pass through the point P and experience a continuous electric field. So, according to the law:
φ = QεO
as we know that the charge would be enclosed inside the Gaussian surface q, and it is equivalent to σ × 4 πR2. Hence,
φ = σ 4πr2εO…….(1)
The overall electric flux that will be passing through the Gaussian surface would be:
φ = E × 4 πr2
or
E = σR2εor2
As the density of surface charge σ is q / 4 πR2.
E = kqr2
- Electric Field Inside the Spherical Shell
To find the electric field inside the spherical shell, we will imagine a point P inside the shell. We could take a spherical Gaussian surface passing through the point P whose centre is O and radius is r. Therefore, according to the law,
φ = qεo=0
So, the total electric flux that is inside the Gaussian surface would be:
φ = E × 4 πr2 = 0
Therefore,
E = 0
Conclusion
Gauss’s law is used to describe the relationship between the distribution of electric charge and electric field. The Electric field is said to be the collection of lines that move in the same direction. The electric field is considered as a vector quantity of two dimensions since it has both magnitude and direction. The concept of the electric field was presented by Michael Faraday. Gauss’s law has many applications and calculating the electric field of a spherical shell is one of them. The produced electric field by the spherical shell can be measured in two ways, i.e. electric field outside the spherical shell and electric field inside the spherical shell.
