Electrostatics Question 253

Question: An infinite number of electric charges each equal to 5 nano-coulomb (magnitude) are placed along X-axis at $ x=1 $ cm, $ x=2 $ cm, $ x=4 $ cm $ x=8 $ cm … and so on. In the setup if the consecutive charges have opposite sign, then the electric field in Newton/Coulomb at $ x=0 $ is $ ( \frac{1}{4\pi {\varepsilon _{0}}}=9\times 10^{9}N-m^{2}/c^{2} ) $ [EAMCET 2003]

Options:

A) $ \frac{Ze^{2}}{2\pi {\varepsilon _{0}}mv^{2}} $

B) $ \frac{Ze}{4\pi {\varepsilon _{0}}mv^{2}} $

C) $ \frac{Ze^{2}}{8\pi {\varepsilon _{0}}mv^{2}} $

D) $ \frac{Ze}{8\pi {\varepsilon _{0}}mv^{2}} $

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Answer:

Correct Answer: A

Solution:

Suppose distance of closest approach is r, and according to energy conservation applied for elementary charge.

Energy at the time of projection = Energy at the distance of closest approach

therefore $ \frac{1}{2}mv^{2}=\frac{1}{4\pi {\varepsilon _{0}}}.\frac{(Ze).e}{r}\Rightarrow r=\frac{Ze^{2}}{2\pi {\varepsilon _{0}}mv^{2}} $



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