Two charged spheres having radii `a` and `b` are joined with a wire, then the ratio of electric fields `E_(a)/E_(b)` on their respective surfaces is

Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surface of two spheres ? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its fatter portions ?

Two spheres of radius a and b respectively are charged and joined by a wire. The ratio of electric field of the spheres is

Two isolated charged conducting spheres of radii a and b produce the same electric field near their surface. The ratio of electric potentials on their surfaces is

Two spheres of radii R_(1) and R_(1) respectively are charged and joined by wire. The ratio of electric field of spheres is

Consider two charged metallic spheres S_1 and S_2 of radii R and 4R, respectively. The electric fields E_1 (on S_1 ) and E_2 (on S_2 ) on their surfaces are such that E_1/E_2 = 5/1 . Then the ratio V_1 (on S_1 ) / V_1 (on S_2 ) of the electrostatic potentials on each sphere is :

Two charged spheres of radii R_(1) and R_(2) having equal surface charge density. The ratio of their potential is

Two isolated metallic spheres of radii 2 cm and 4 cm are given equal charge, then the ratio of charge density on the surfaces of the spheres will be

Two spheres of same metal have radii a and b . They have been connected to a conducting wire. Find the ratio of the electric field intensity upon them.

Two spherical conductors A and B of radii 1mm and 2mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is

Two conducting spheres of radii R_(1) and R_(2) are at the same potential. The electric intensities on their surfaces are in the ratio of