Question:
A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 × 105 NC–1. If the charge on the particle is 40 μC and the initial velocity is 200 ms–1, how much distance it will travel before coming to the rest momentarily?
A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 × 105 NC–1. If the charge on the particle is 40 μC and the initial velocity is 200 ms–1, how much distance it will travel before coming to the rest momentarily?
Updated On: Mar 19, 2025
- 1 m
- 5 m
- 10 m
- 0.5 m
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The Correct Option is D
Solution and Explanation
The correct option is(D): 0.5 m
\(v^2-u^2=2aS\)
\(⇒s=\frac{4}{2x4}m=0.5m\)
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).