An electron falls through a distance of 1.5, cm in a uniform electric field of magnitude 2.0×10^4,

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1. Electric Charges and Fields

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An electron falls through a distance of $1.5\, cm$ in a uniform electric field of magnitude $2.0\times10^4\, N/C$ as shown in the figure. The time taken by electron to fall through this distance is ($m_e = 9.1\times10^{-31}\,kg$, Neglect gravity)

A

$1\times10^{-9}\,s$

B

$2.9\times10^{-9}\,s$

C

$2.9\times10^{-8}\,s$

D

$1\times10^{-8}\,s$

Solution

Electric force $= qE = 1.6 \times {10^{ – 19}} \times 2 \times {10^4} = 3.2 \times {10^{ – 15}}N$

$acc$ of ${e^ – } = \frac{{3.2 \times {{10}^{ – 15}}}}{{9.1 \times {{10}^{ – 31}}}}$

$S = \frac{1}{2}a{t^2}$

$t = \sqrt {\frac{{2S}}{a}} = \sqrt {\frac{{2 \times 1.5 \times {{10}^{ – 2}} \times 9.1 \times {{10}^{ – 31}}}}{{3.2 \times {{10}^{ – 15}}}}} $

$=2.9\times10^{-9}\,s$

Standard 12

Physics

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