Q. A spherical charged conductor has surface density of charge$= \sigma$, and electric field intensity on its surface is E. If radius of surface is doubled, point $\sigma$ unchanged, what will be electric field intensity on the new sphere?
A) E/2
B) 2 E
C) E/4
D) E
Explanation
(d) $E = \sigma/\varepsilon_{0},$ The value of $E$ does not depend upon radius of the sphere.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.