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Electric field intensity due to infinite sheet of charge

Consider a sheet of infinite extent on which charges are uniformly distributed with charge density σ. We want to calculate electric intensity E at point P at a distance r from the sheet. Imagine a Gaussian surface in the form of a cylinder passing through the sheet. Let the height of the cylinder is 2r and cross-sectional area A. As E is perpendicular to the end faces and parallel (angle of E and area of the curved surface is 90°.) to the curved faces of the cylinder, therefore, the flux through the curved faces is zero and all flux contribution come through the end faces.

Since charge per unit area is σ and area enclosed is ΔA, therefore, total charge enclosed by the Gaussian surface is Q = σ ΔA. Put this value in the above equation.

In vector form

where n is a unit vector normal to the sheet.

Electric Field intensity due to a charge conducting spherical shell

Consider a charged conducting spherical shell (a hollow one) of radius R. Suppose a +ve charge Q is given to the shell which distributes evenly on the surface. We calculate E on point p inside the sphere (rB ˂ R).

Consider a point B situated inside the shell at a distance r from the center of the sphere, hence rB is less than R . Apply Gauss’s law by taking a Gaussian surface of radius rB through the point B. As there is no charge inside the Gaussian surface, therefore, Q=0 and,

The charge is evenly distributed on the surface, therefore,(1) E is constant at all points at radial distance rB from the centre (think over, E will be constant at the surface of any sphere drawn inside this sphere). (2) Vector ΔA̅ is also radial, therefore, both E̅ and ΔA̅ have same direction, therefore,

Sum of ΔA is the surface area and equal to 4πrB2, therefore, E (4πrB2) = 0

Or E = 0    (as 4πrB2≠ 0)

B was chosen any arbitrary point; therefore, we conclude that the field inside a charged spherical shell is zero.

(How can you explain the safety of the passengers of airplanes in thunderstorms).

You can calculate the field intensity when the point p is outside the charge sphere. It will be

This result is identical to Coulomb’s law. Do it yourself.

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