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Question

Considering gaussian surface

$\phi=\frac{q_{	ext {in }}}{\varepsilon_{0}} \Rightarrow E 4 \pi d^{2}=\frac{2 e d^{3}}{\varepsilon_{0} R^{3}}$

$\

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$\frac{R}{2}$

Vedclass · Answer

Considering gaussian surface

$\phi=\frac{q_{	ext {in }}}{\varepsilon_{0}} \Rightarrow E 4 \pi d^{2}=\frac{2 e d^{3}}{\varepsilon_{0} R^{3}}$

$\mathrm{E}=\frac{2 \mathrm{edk}}{\mathrm{R}^{3}}$

For equilibrium of charge,

$\frac{	ext { kee }}{(2 d)^{2}}=\left(\frac{	ext { k } 2 	ext { ed }}{R^{3}}ight) e$

$\frac{1}{4 \mathrm{d}^{2}}=\frac{2 \mathrm{d}}{\mathrm{R}^{3}} \Rightarrow \mathrm{d}^{3}=\frac{\mathrm{R}^{3}}{8} \Rightarrow \mathrm{d}=\mathrm{R} / 2$

Using thomson's model of the atom, consider an atom consisting of two electrons, each of charge $-e$, embeded in a sphere of charge $+2e$ and radius $R$. In equilibrium each electron is at a distance $d$ from the centre of the atom. What is the equilibrium separation between electrons

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