Electric field intensity due to charged disc

A charge Q is placed at the centre of a square If electric field intensity due to the charge at the corners of the square is E_(1) and the intensity at the mid point of the sidde of square is E_(2) then the ratio of (E_(1))/(E_(2)) will be

A point charge of 100mu C is placed at 3 hat i+4 hat j m . Find the electric field intensity due to this charges at a point located at 9 hat i+12 hat j m .

An uncharged metallic solid sphere of radius R is placed at a distance 2R from point charge Q as shown in figure. The electric field intensity due to induced charge at centre of sphere is

Electric field intensity (E) due to an electric dipole varies with distance (r ) from the point of the center of dipole as :

Electric field intensity at a point due to an infinite sheet of charge having surface charge density sigma is E .If sheet were conducting electric intensity would be

Find the electric field on the axis of a charged disc.

The figure shown is a plot of electric field lines due to two charges Q_(1) and Q_(2) . The sign of charges is

A spherical charged conductor has surface density of charge as sigma . The electric field intensity on its surface is E . If the radius of the surface is doubled, keeping sigma uncharged, what will be the electric field intensity on the new sphere ?

The electric field intensity vec(E) , , due to an electric dipole of dipole moment vec(p) , at a point on the equatorial line is :

At s point due to a point charge, the values of electric field intensity and potential are 32 NC^-1 and 16 JC^-1 , respectively. Calculate the a. magnitude of the charge, and b. distance of the charge from the point of observation.