Derive expression for the electric field ‘just’ outside a charged ‘conductor’.

State Gauss’ theorem in electrostatics.
Apply this theorem to derive an expression for the electric intensity due to an infinite sheet of charge and hence deduce the expression for electric intensity due to two parallel oppositely charged thin infinite sheets of charge at a point

(ii) Outside the sheets.

Using Gauss’ theorem derive an expression for the electric intensity due to a charged thin spherical shell at

(i) external point,

(ii) surface point,

(iii) internal point.
Draw the graph showing the variation of electric intensity obtained as above with distance from the centre.

Using Gauss’ theorem derive an expression for the electric intensity due to a charged thin spherical shell at

internal point.

Draw the graph showing the variation of electric intensity obtained as above with distance from the centre.

A point-charge produces an electric flux of $-750 \mathrm{N} {\mathrm{m}}^2 {\mathrm{C}}^{-1}$ through a spherical Gaussian surface of $10 \mathrm{cm}$radius with centre at the charge,

 If the radius of the Gaussian surface be doubled, how much flux would pass through the surface?