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}} $
Show Answer
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}} $
