Electric Field Due To A Line Charge

• Electric Field Due To A Line Charge

Consider a thin infinitely long straight line charge of linear charge density λ.
Let P be the point at a distance ‘a’ from the line. To find electric field at point P, draw a cylindrical surface of radius ‘a’ and length l.
If E is the magnitude of electric field at point P, then electric flux through the Gaussian surface is given by,

Φ = E × Area of the curved surface of a cylinder of radius r and length l

Because electric lines of force are parallel to end faces (circular caps) of the cylinder, there is no component of field along the normal to the end faces.

Φ = E × 2πal … (i)
According to Gauss theorem, we have

From equations (i) and (ii), we obtain

• Electric Field Due To An Infinite Plane Sheet Of Charge
  

Consider an infinite thin plane sheet of positive charge having a uniform surface charge density σon both sides of the sheet. Let P be the point at a distance ‘a’ from the sheet at which electric field is required. Draw a Gaussian cylinder of area of cross-section A through point P.
The electric flux crossing through the Gaussian surface is given by,

Φ = E × Area of the circular caps of the cylinder

Since electric lines of force are parallel to the curved surface of the cylinder, the flux due to electric field of the plane sheet of charge passes only through the two circular caps of the cylinder.

Φ = E × 2A … (i)
According to Gauss theorem, we have

Here, the charge enclosed by the Gaussian surface,
q = σA

From equations (i) and (ii), we obtain

• Electric Field Due To A Uniformly Charged Thin Spherical Shell
  

• When point P lies outside the spherical shell
  

Suppose that we have to calculate electric field at the point P at a distance r (r > R) from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius r and centre O.
Let be the electric field at point P. Then, the electric flux through area element is given by,

Since is also along normal to the surface,
dΦ = E ds
∴ Total electric flux through the Gaussian surface is given by,

Now,

Since the charge enclosed by the Gaussian surface is q, according to Gauss theorem,

From equations (i) and (ii), we obtain

• When point P lies inside the spherical shell
  

In such a case, the Gaussian surface encloses no charge.
According to Gauss law,
E × 4πr² = 0
i.e., = E = 0 (r < R)

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