A large nonconducting surface has a uniform charge density `sigma`. A small circular hole ofradius R is cut in the middle of the sheet, as shown in figure. Ignore fringing of the field lines around all edges calculate the electric field at' pointP, a distancez from the centre of the hole along its axis.

The surface charge density of a thin charged disc of radius R is sigma . The value of the electric field at the centre of the disc is (sigma)/(2in_(o)) . With respect to the field at the centre, the electric field along the axis at a distance R from the centre of the disc reduces by

A circular ring carries a uniformly distributed positive charge .The electric field(E) and potential (V) varies with distance (r ) from the centre of thr ring along its axis as

Consider a thin spherical shell of radius R consisting of uniform surface charge density sigma . The electric field at a point of distance x from its centre and outside the shell is

An infinitely large non-conducting plane of uniform surface charge density sigma has circular aperture of certain radius carved out from it. The electric field at a point which is at a distance 'a' from the centre of the aperture (perpendicular to the plane) is (sigma)/(2 sqrt(2) in_(0)) . Find the radius of aperture :

Three non-conducting large parallel plates have surface charge densities sigma, -2 sigma and 4 sigma respectively as shown in the figure. The electric field at the point P is

A non-conducting semi circular disc (as shown in figure) has a uniform surface charge density sigma . The ratio of electric field to electric potential at the centre of the disc will be

.A thin disc of radius b=2a has a concentric hole of radius a' in it.It carries uniform charge density sigma on it.If the electric field on its axis at distance 'd' from the centre is given as 'nd' then value of 'n' is

In the figure shown S is a large nonconducting sheet of uniform charge density sigma . A rod R of length l and mass 'm' is parallel to the sheet and hinged at its mid point. The linear charge densities on the upper and lower half of the rod are shown in the figure. Find the angular acceleration of the rod just after it is released.