How long after the particle is released has it travelled through a potential dif

Question

How long after the particle is released has it travelled through a potential difference of 0.601 volts?

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A particle with charge-6.74 x 10-6 coulombs is released at rest in a region of constant, uniform electric field. Assume that gravitational effects are negligible. At a time 7.36 seconds after it is released, the particle has a kinetic energy of 6.13 x 10a joules. How long after the particle is released has it travelled through a potential difference of 0.601 volts? Number

Explanation / Answer

at t = 7.36 sec ,

KE = 6.13 x 10^-8 = (m v^2 ) /2

v = sqrt(12.26 x 10^-8 / m )

and suppose field strength is E.

then a = qE/m

so from 0 to 7.36 sec

vf = vi + at

sqrt(12.26 x 10^-8 / m ) = 0 + (qE/m)t

sqrt(12.26 x 10^-8 / m ) = a 7.36

a =4.76 x 10^-5 / sqrt(m)

when particle traveled through 0.601 volts
gain in KE = qV = 6.74 x 10^-6 x 0.601 = 4.05 x 10^-6 J

mv^2 /2 = 4.05 x 10^-6

v = 2.85 x 10^-3 / sqrt(m)

using vf = vi + at

2.85 x 10^-3 / sqrt(m) = (4.76 x 10^-5 / sqrt(m) ) t

t = 59.87 sec

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