Question

An electron falls through a distance of 1.5 cm in electric field of magnitude 2×10×10×10×10 n/c .Now direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance . Compute the time of fall in each case neglecting gravity

Answer 1

Answer: $t_{e}= 2.9\times 10^{-9} s$ and $t_{p} = 1.2\times10^{-7}\ s$

Explanation:

Given that,

Distance $s = 1.5 cm$

Electric field $E = 2\times10^{4} N/C$

Mass of electron $M_{e} = 9.1\times 10^{-31}\ kg$

Mass of proton $M_{p} = 1.6\times 10^{-27}\ kg$

Charge $q = 1.6\times 10^{-19}\ C$

Distance travelled by both particle

Now, using of equation of motion

$s = ut +\dfrac{1}{2} at^{2}$

Initial velocity u = 0 m/s

So, $s = \dfrac{1}{2} at^{2}$

$t = \sqrt{\dfrac{2s}{a}$.....(I)

We know that,

The electric force is

F = qE

Now, newton's second law

F = ma

Now, on equating to  both force

F = qe = ma

So, $a = \dfrac{qE}{m}$....(II)

Now, put the value of a in equation (I)

The time of fall in each case

$t = \sqrt{\dfrac{2sm}{qE}$...(III)

Now, put the value of all elements in equation (III)

For electron,

$t = \sqrt{\dfrac{2\times 1.5\times 10^{-2}\ m\times 9.1\times 10^{-31}\ kg}{1.6\times 10^{-19}\ C \times 2\times 10^{4}\ N/C}$

$t_{e}= 2.9\times 10^{-9} s$

Now, for proton

$t = \sqrt{\dfrac{2\times 1.5\times 10^{-2}\ m\times 1.6\times 10^{-27}\ kg}{1.6\times 10^{-19}\ C \times 2\times 10^{4}\ N/C}$

$t_{p} = 1.2\times10^{-7}\ s$

Hence, this is the required solution.

Answer 2

Answer:

Explanation:

Answer:  and

Explanation:

Given that,

Distance

Electric field

Mass of electron

Mass of proton

Charge

Distance travelled by both particle

Now, using of equation of motion

Initial velocity u = 0 m/s

So,

.....(I)

We know that,

The electric force is

F = qE

Now, newton's second law

F = ma

Now, on equating to  both force

F = qe = ma

So, ....(II)

Now, put the value of a in equation (I)

The time of fall in each case

...(III)

Now, put the value of all elements in equation (III)

For electron,

Now, for proton

Hence, this is the required solution.

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