બે ઈલેક્ટ્રોનને '2d' અંતરે જડિત | vedclass

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Question

From the given condition, we have

$F _{	ext {net}}=-\left[2 F _{ q / q } \cos 	hetaight]$

$F _{	ext {net}}=-2 \cdot \frac{1}{4 \pi \varepsi

Vedclass · Accepted Answer

$\left(\frac{ q ^{2}}{2 \pi \varepsilon_{0} md ^{3}}\right)^{\frac{1}{2}}$

Vedclass · Answer

From the given condition, we have

$F _{	ext {net}}=-\left[2 F _{ q / q } \cos 	hetaight]$

$F _{	ext {net}}=-2 \cdot \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{ q ^{2}}{\left(\sqrt{ d ^{2}+ x ^{2}}ight)^{2}} \cdot \frac{ x }{\sqrt{ d ^{2}+ x ^{2}}}$

$=-\frac{q^{2}}{2 \pi \varepsilon_{0}} \frac{x}{\left(d^{2}+x^{2}ight)^{3 / 2}}$

For $x ( x << d )$

$F_{	ext {net }}=-\frac{q^{2}}{2 \pi \varepsilon_{0} d^{3}} x$

$	herefore a =-\frac{ q ^{2}}{2 \pi \varepsilon_{0} \cdot md ^{3}} x$

Comparing with equation of $SHM \left( a =-\omega^{2} x ight)$

$	herefore \omega=\sqrt{\frac{ q ^{2}}{2 \pi \varepsilon_{0} md ^{3}}}$

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