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)at^{2}

here

initial velocity, u = 0 m/s

so,

s = (1/2)at^{2}

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 × 10^{4} N/C

Now, for an electron

m = 9.1 x 10^{-31} kg

q = 1.6 x 10^{-19} C

thus, we get

t_{electron} = [ (2 x 9.1x10^{-31 }x 0.015) / (1.6x10^{-19} x 2.0 × 10^{4})]^{1/2}

or

**t**_{electron}** = 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

t_{proton} = [ (2 x 1.6x10^{-27 }x 0.015) / (1.6x10^{-19} x 2.0 × 10^{4})]^{1/2}

or

**t**_{proton}** = 1.3 x 10**^{-7}** s**

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