JEE Main 2022 Physics question
A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength . If the charge on the particle is 40 and the initial velocity is 200 , how much distance it will travel before coming to the rest momentarily?
Answer: (4)
Solution
The deceleration is a = qE/m = 10^{-6}10^{5})10^{-6})10^{4}$
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