# An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 10 4  N C –1   in the direction of the field . Then the  direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance . compute the time of fall in each case.

distance travelled by each particle will be given as

s = ut + (1/2)at2

here

initial velocity, u = 0 m/s

so,

s = (1/2)at2

or

time taken

t = (2s/a)1/2 ..................(1)

now,

the force exerted by the electric field causes acceleration in the particle, so

F = qE = ma

or

acceleration

a = qE/m .................(2)

thus,

equation (1) becomes

t = (2s / qE/m)1/2

or

t = (2ms / qE)1/2 ...................(3)

we shall use this equation to calculate the time taken by each particle

.

.

Times of fall

given

s = 1.5cm = 0.015m

E = 2.0 × 104  N/C

Now, for an electron

m = 9.1 x 10-31 kg

q = 1.6 x 10-19 C

thus, we get

telectron = [ (2 x 9.1x10-31 x 0.015) / (1.6x10-19 x  2.0 × 104)]1/2

or

telectron =  2.9 x 10-9 s

and

For a proton

m = 1.6 x 10-27 kg

q = 1.6 x 10-19 C

thus, we get

tproton = [ (2 x 1.6x10-27 x 0.015) / (1.6x10-19 x  2.0 × 104)]1/2

or

tproton =  1.3 x 10-7 s

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