Electrostatics Question 236
Question: A charge $ +q $ is fixed at each of the points $ x=x _{0},x=3x _{0},x=5x _{0} $ ….. infinite, on the $ x- $ axis and a charge $ -q $ is fixed at each of the points $ x=2x _{0},x=4x _{0},x=6x _{0} $ ,….. infinite. Here $ x _{0} $ is a positive constant. Take the electric potential at a point due to a charge $ Q $ at a distance $ r $ from it to be $ Q/(4\pi {\varepsilon _{0}}r) $ . Then, the potential at the origin due to the above system of charges is [IIT 1998]
Options:
A) Electric field near $ A $ in the cavity = Electric field near $ B $ in the cavity
B) Charge density at $ A $ equals charge density at $ B $
C) Potential at $ A $ equals Potential at $ B $
D) Total electric field flux through the surface of the cavity is $ q/{\varepsilon _{0}} $
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Answer:
Correct Answer: C , D
Solution:
Under electrostatic condition, all points lying on the conductor are in same potential. Therefore, potential at A = potential at B.
From Gauss’s theorem, total flux through the surface of the cavity will be $ q/\varepsilon_{0} $ .
Not: q Instead of an elliptical cavity, if it had been a spherical cavity then options (a) and (b) would also have been correct.
Under electrostatic condition, all points lying on the conductor are in same potential. Therefore, potential at A = potential at B.
From Gauss’s theorem, total flux through the surface of the cavity will be $ q/\varepsilon_{0} $ .
BETA
