Saturday, August 3, 2019

Electric Fields: Exercise 1

Calculate the speed of an electron accelerated from rest through a distance of 40 mm by a uniform electric field of 3.0 × 103 N C−1.

Solution:


force = Eq = (3 × 103 × 1.6 x 10−19) = 4.8 × 10−16 N
acceleration = force/mass = 4.8 × 10−16 /9.1 × 10−31 = 5.27 × 1014 ms−2
v2 = 0 + 2as = 2 × 5.27 × 1014 ms−2 × 40 × 10−3 so speed v = 6.5 × 106 m s−1

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