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An electron falls through a distance of $1.5\; cm$ in a uniform electric field of magnitude $2.0\times 10^{4}\; N/C$ (Fig. a)

Calculate the time it takes to fall through this distance starting from rest.

If the direction of the field is reversed (fig. b) keeping its magnitude unchanged, calculate the time taken by a proton to fall through this distance starting from rest.

Answers (1)

$\left | \vec{E} \right |=2\times 10^{4}\; N/C$

$d=1.5 \; cm$

$a=\frac{eE}{m}=\frac{1.6\times 10^{-19}\times 2\times 10^{4}}{9.31\times 10^{-31}}=0.34\times 10^{16}\; m/s^{2}$

$S=ut+\frac{1}{2}at^{2}$

$S=\frac{1}{2}at^{2}$

$t=\sqrt{\frac{2S}{a}}$

$=\sqrt{\frac{3\times 10^{-2}}{0.34\times 10^{16}}}$

$=\sqrt{8.82\times 10^{-18}}$

$=2.96\times 10^{-9}\; sec$

$=2.96\; ns$

For proton mass $=1.67\times 10^{-27}$

$\therefore a=\frac{1.69\times 10^{-19}\times 2\times 10^{4}}{1.67\times 10^{-27}}\approx 2\times 10^{12}m/s^{2}$

$t=\sqrt{\frac{3\times 10^{-2}}{2\times 10^{12}}}=\sqrt{1.5\times 10^{-14}}$

$=1.2\times 10^{-7}\; sec$

$=120 \; n\; sec$

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Safeer PP

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