Question

An electron initially at rest falls a distance of 2 cm in a
uniform electric field of magnitude 3 x 104 N 0-1. The
time taken by the electron to fall this distance is
a)1.3×10^-12s
b)1.6×10^-10s
c)2.1×10^-12s
d)2.75×10^-9s​

Answer 1

Answer:

$force \: experienced \: by \: electron \\ of \: mass \: m \: and \: charge \: e$

$f = m.a$

$the \: acceleration \: of \: electron \\ is$

a = e/m

$now \: equations \: of \: motion$

$v = u + at$

$and$

$h = ut + \frac{1}{2} a {t}^{2}$

$as \: the \: body \: is \: starting \: from \\ rest \: u = 0 \: the \: time \: required \\ by \: the \: electron \: to \: fall \: through \\ a \: distance \: h \: is \: given \: as:$

$t = \sqrt{ \frac{2h}{a} } = \sqrt{ \frac{2hm}{e} }$

$e = 1.6 \times {10}^ - 19{}$

$m = 9.11 \times {10}^ - 31{}$

$e = 3 \times {10}^4{}$

$h = 2 \times {10}^ - 2{}$

$t = \sqrt{ \frac{3 \times 2 \times {10}^ - 2{m \times 9.11 \times {10}^ - 31{} }kg }{1.6 \times {10}^ - 19{}c \times 2.0 \times {10}^4{} } }$

$t = 2.75 \times {10}^ - 9{} s$

Explanation:

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Answer 2

Answer:

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