Electrostatics Question 785
Question: A charge +q fixed at each of the points $ x=x _{0},x=3x _{0},x=5x _{0}…. $ up to $ \infty $ on X-axis and charge -q is fixed on each of the points $ x=2x _{0},x=4x _{0},…. $ up to $ \infty $ . Here $ x _{0} $ is a positive constant. Take the potential at a point due to a charge Q at a distance r form it to be $ \frac{Q}{4\pi {\varepsilon _{0}}r}, $ then the potential at the origin due to above system of charges will be:
Options:
A) zero
B) infinite
C) $ \frac{q}{8\pi {\varepsilon _{0}}x _{0}{\log _{e}}2} $
D) $ \frac{q{\log _{e}}2}{4\pi {\varepsilon _{0}}x _{0}} $
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Answer:
Correct Answer: D
Solution:
[d] $ V=\frac{1}{4\pi {\varepsilon _{0}}}[ \frac{q}{x _{0}}+\frac{q}{3x _{0}}+\frac{q}{5x _{0}}+… ] $
