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(A) According to Thomson's model,the positive charge $+2e$ is uniformly distributed in a sphere of radius $R$. The charge density is $\rho = \frac{2e}

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$R$

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(A) According to Thomson's model,the positive charge $+2e$ is uniformly distributed in a sphere of radius $R$. The charge density is $ho = \frac{2e}{\frac{4}{3}\pi R^3} = \frac{6e}{4\pi R^3}$.For an electron at distance $d$ from the center,the electric field $E$ due to the positive charge sphere is found using Gauss's Law:$E(4\pi d^2) = \frac{q_{	ext{in}}}{\varepsilon_0} = \frac{ho (\frac{4}{3}\pi d^3)}{\varepsilon_0} = \frac{(\frac{6e}{4\pi R^3})(\frac{4}{3}\pi d^3)}{\varepsilon_0} = \frac{2ed^3}{\varepsilon_0 R^3}$.Thus,$E = \frac{2ed}{4\pi\varepsilon_0 R^3} = \frac{2ked}{R^3}$,where $k = \frac{1}{4\pi\varepsilon_0}$.For equilibrium,the electrostatic force of repulsion between the two electrons must be balanced by the attractive force from the positive sphere:$F_{	ext{repulsion}} = F_{	ext{attraction}}$$\frac{ke^2}{(2d)^2} = eE = e \left( \frac{2ked}{R^3} ight)$$\frac{ke^2}{4d^2} = \frac{2ke^2d}{R^3}$$\frac{1}{4d^2} = \frac{2d}{R^3} \Rightarrow 8d^3 = R^3 \Rightarrow d^3 = \frac{R^3}{8} \Rightarrow d = \frac{R}{2}$.The separation between the two electrons is $2d = 2(R/2) = R$.

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

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