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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, 10:54: PM
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, 02:45: PM
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