Case I - Uniform Field
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The figure shows field lines passing through a rectangular surface of area A perpendicular to the field lines.
The electric flux passing through this surface is given by the product of electric intensity and the surface area perpendicular to t
he field lines.
f = EA where f denotes electric flux and A denotes the surface area.
Suppose the surface is not perpendicular to the field lines, then the electric flux is give
n by the equation
f = EA cos q
where
q is the angle between the direction of electric field E and the normal drawn to the surface in the outward direction.
When the normal to surface is parallel to the electric field as shown in the figure, the electric flux is
f = EA Cos q, (q = 0)
f = EA Cos 0
f = EA
The electric flux becomes zero if the normal to the surface is perpendicular to the electric field as shown in the figure.
That is,
f = EA Cos
q, (
q = 90
o)
= EA Cos 90
= EA x 0
f = 0
Case II - Non - uniform Field
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Let us now calculate the electric flux passing through a surface when the applied field is not uniform. The surface is usually divided into a large number of small area dA such that the electric field remains constant over that surface as shown in the figure.
The electric flux passing through dA(df) = EdA cosq
S.I unit of electric flux is Nm
2/c
2
The electric flux per unit area is defined as the electric flux density.
Sample Problems to calculate Electric Flux
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01. A point charge of 1.8mC is at the centre of a cubical Gaussain surface having each side 50cm. What is the net electric flux through the surface?
Suggested answer:
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02. An electric dipole consists of +5C and -5C separated by a distance 3cm. Find the flux passing through a sphere of radius 6cm enclosing the dipole.
Suggested answer:
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