A particle with a mass of 2.0 x10^-5 kg and a charge of +2.0µC is released in a (parallel plate) uniform...?
A particle with a mass of 2.0 x10^-5 kg and a charge of +2.0µC is released in a (parallel plate) uniform horizontal electric field of 12 N/C.
How far horizontally does the particle travel in 0.50s?
Answer is x = 0.15m
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By F = qE
=>ma = qE
=>a = qE/m
=>a = [2 x 10^-6 x 12]/[2 x 10^-5]
=>a = 1.2 m/s^2
Thus by s = ut + 1/2at^2
=>s = 0 + 1/2 x 1.2 x (0.5)^2
=>s = 0.15m OR 15 cm