SATHEE: Electrostatics Question 253

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 $(\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N - m^{2} / c^{2})$ [EAMCET 2003]

Options:

A) $\frac{Z e^{2}}{2 π ε_{0} m v^{2}}$

B) $\frac{Z e}{4 π ε_{0} m v^{2}}$

C) $\frac{Z e^{2}}{8 π ε_{0} m v^{2}}$

D) $\frac{Z e}{8 π ε_{0} m v^{2}}$

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 $\frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} . \frac{(Z e) . e}{r} \Rightarrow r = \frac{Z e^{2}}{2 π ε_{0} m v^{2}}$

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Electrostatics Question 253

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