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

Two charged conducting spheres of radill a and b are connected to eachother 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 conducter 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 conducting spheres of radii r_(1) and r_(2) are charged to the same surface charge density . The ratio of electric field near their surface 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

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 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.