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 radio of the magnitude of the electric fields at the surfaces of spheres A and B is
A) 4 : 1
B) 1 : 2
C) 2 : 1
D) 1 : 4
Explanation
(c) Here $\frac{r_{1}}{r_{2}} = \frac{1mm}{2mm} = \frac{1}{2}.$ when the spheres are connected by a conducting wire $V_{1} = V_{2}$
Or $\frac{q_{1}}{4\pi\varepsilon_{0}r_{1}} = \frac{q_{2}}{4\pi\varepsilon_{0}r_{2}} \Rightarrow \frac{q_{1}}{q_{2}} = \frac{r_{1}}{r_{2}} = \frac{1}{2}$
Now, $\frac{E_{1}}{E_{2}} = \frac{q_{1}}{q_{2}}.\left( \frac{r_{2}}{r_{1}} \right)^{2} = \frac{1}{2} \times \left( \frac{2}{1} \right)^{2} = 2\ :1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.