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An electron is projected with velocity 10^7m/s at an angle  A=30 degrees with horizontal in a region of uniform electric field  of 500N/C vertically upwards.Find the maximum distance covered by an electron in vertical direction above its initial  level.
Asked by jmorakhia | 26 Apr, 2019, 22:54: PM
answered-by-expert Expert Answer
This problem is similar to projectile motion in gravitational field.
 
downward acceleration a of electron in electric field E is given by, 
 
a =   e E/m  = (e/m)E  = 1.759×1011 × 500   = 8.795×1013 m/s2
 
where e is charge on electron, m is mass of electron.
 
hence vertical distance , h  = u2 sin2θ / a  = 1014 /(4×8.795×1013) ≈ 0.28 m  or 28 cm
Answered by Thiyagarajan K | 27 Apr, 2019, 14:45: PM

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