Question

.An electron initially at rest falls a distance of 1.5 cm
in a uniform electric field of magnitude 2 x 104 N/C.
The time taken by the electron to fall this distance is

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Answer 1

$\underline{ \boxed{ \bold{ \mathfrak{ \huge{ \purple{Answer}}}}}}$

Given :

distance travelled= 0.015 m

electric field intensity = 2×10^4 N/C

initial velocity = 0 m/s

To Find :

Time taken by the electron to fall this distance is..

Formula :

$\dag \rm \: \blue{net \: force = electrical \: force} \\ \\ \therefore \rm \: ma = qE \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \pink{a = \frac{qE}{m} }}}}} \\ \\ \dag \rm \: kinematic \: equation... \\ \\ \underline{ \boxed{ \bold{ \rm{ \blue{d = ut + \frac{1}{2} a {t}^{2} }}}}} \\ \\ \dag \rm \: mass \: of \: electron = 9.109 \times {10}^{ - 31} \: kg$

Calculation :

$\rm :\implies \: a = \frac{qE}{m} = \frac{1.6 \times {10}^{ - 19} \times 2 \times {10}^{4} }{9.109 \times {10}^{ - 31} } \\ \\ \therefore \: \underline{ \rm\boxed{\red{a = 3.5 \times {10}^{16} \: \frac{m}{ {s}^{2} } }}} \\ \\ : \implies \rm \: d = \frac{1}{2} a {t}^{2} \: \: \: \: ( \because \: u = 0) \\ \\ \therefore \rm \: 0.015 = \frac{1}{2} \times 3.5 \times {10}^{16} \times {t}^{2} \\ \\ \therefore \rm \: t = \sqrt{ \frac{0.015 \times 2}{3.5 \times {10}^{16} } } = \sqrt{8.57 \times {10}^{ - 19} } \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{t = 0.92 \: nS}}}}}$