Q. 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
A) 1:2
B) 2:1
C) 1:4
D) 4:1
Explanation
(b) After connecting through a wire $V_{A} = V_{B}$
$\Rightarrow \frac{kQ_{A}}{r_{A}} = \frac{kQ_{B}}{r_{B}} \Rightarrow \frac{Q_{A}}{Q_{B}} = \frac{r_{A}}{r_{B}}$
Ratio of electric field
$\frac{E_{A}}{E_{B}} = \frac{Q_{A}}{Q_{B}} \times \left( \frac{r_{B}}{r_{A}} \right)^{2}\ \ \ \ \left\lbrack \because E = \frac{kQ}{r^{2}} \right\rbrack$
$\Rightarrow \frac{E_{A}}{E_{B}} = \frac{r_{A}}{r_{B}} \times \left( \frac{r_{B}}{r_{A}} \right)^{2} = \frac{r_{B}}{r_{A}} = \frac{2}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.