[" A uniform electric field of "],[" magnitude "5.8times10^(2)N/" passes "],[" through a circular disc of radius "],[" 10cm.Ratio of electric flux "],[" through the disc when its face is "],[" perpendicular to the field lines and "],[" when its face is at "45^(@)" to the field "],[" lines,is "]

[" A uniform electric field of "],[" magnitude "5.8times10^(2)N/" C passes "],[" through a circular disc of radius "],[10cm" .Ratio of electric flux "],[" through the disc when its face is "],[" perpendicular to the field lines "],[" and when its face is at "45^(@)" to the "],[" field lines,is "]

[" A uniform electric field of "],[" magnitude "5.8times10^(2)N/" C passes "],[" through a circular disc of radius "],[10cm." Ratio of electric flux "],[" through the disc when its face is "],[" perpendicular to the field lines "],[" and when its face is at "45^(@)" to the "],[" field lines,is "]

A uniform electric field of magnitude 5.8times10^(2)N/C passes through a circular disc of radius "10cm" . Ratio of electric flux through the disc when its face is perpendicular to the field lines and when its face is at 45^(@) to the field lines,is (1) (1)/(sqrt(2)) (2) 1 (3) sqrt(2) (4) 2

Calculate the moment of inertia of uniform circular disc of mass 500 g, radius 10 cm about : the axis through the centre of the disc and perpendicular to its plane

The flux of electric field through the circular region of radius r as shown in the figure is

What is electriG flux ? Explain how the electric flux through a surface is related to electric field intensity, when the surface is heldinside the electric field.

An electric dipole free to move is placed in a uniform electric field. Explain with a diagram. Its motion, when it is placed perpendicular to the field.

The electric flux through a hemispherical surface of radius R placed in a uniform electric field E parallel to the axis of the circular plane is