JEE MAIN - Physics (2006 - No. 33)
Two spherical conductors $$A$$ and $$B$$ of radii $$1$$ $$mm$$ and $$2$$ $$mm$$ 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
$$4:1$$
$$1:2$$
$$2:1$$
$$1:4$$
स्पष्टीकरण
After connection, $${V_1} = {V_2}$$
$$ \Rightarrow K{{{Q_1}} \over {{r_1}}} = K{{{Q_2}} \over {{r^2}}}$$
$$ \Rightarrow {{Q{}_1} \over {{r_1}}} = {{{Q_2}} \over {{r_2}}}$$
The ratio of electric fields
$${{{E_1}} \over {{E_2}}} = {{K{{{Q_1}} \over {r_1^2}}} \over {K{{{Q_2}} \over {r_2^2}}}} = {{{Q_1}} \over {r_1^2}} \times {{r_2^2} \over {{Q_2}}}$$
$$ \Rightarrow {{{E_1}} \over {{E_2}}} = {{{r_1} \times r_2^2} \over {r_1^2 \times {r_2}}} \Rightarrow {{{E_1}} \over {{E_2}}} = {{{r_2}} \over {{r_1}}} = {2 \over 1}$$
Since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.