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प्रश्न
A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. How much distance will it travel before coming to momentary rest ?
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उत्तर
Given:
Charge of the particle = q
Velocity of projection = u
Electric field intensity = E
Mass of the particle = m
We know that the force experience by a charged particle in an electric field is qE.
Acceleration produced, a = \[\frac{qE}{m}\] (Negative because the particle is thrown against the electric field)
Let the distance covered by the particle be s.
Then v2 = u2 + 2as
[Here, a = deceleration, v = final velocity]
Here, 0 = u2 − 2as
\[\Rightarrow s = \frac{u^2}{2a}\]
\[ \Rightarrow s = \frac{m u^2}{2qE}\]
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