# A conductive sphere of radius r is given a charge Q. Another sphere of radius R is put in contact with it.They are separated after they have shared a charge. assuming that the spheres are held at fixed distance apart, what should be the value of R so that the force between two charged spheres is the maximum?..if u can help me out!

Dear Student,

Please find below the solution to the asked query:

Potential of the first sphere and second sphere after the charge distribution are,

${V}_{1}=\frac{1}{4\pi {\epsilon}_{0}}\frac{{Q}_{1}}{r}and{V}_{2}=\frac{1}{4\pi {\epsilon}_{0}}\frac{{Q}_{2}}{R}$

When these two spheres are kept in contact, then the potential of the two spheres should be same. Therefore,

${V}_{1}={V}_{2}\Rightarrow \frac{1}{4\pi {\epsilon}_{0}}\frac{{Q}_{1}}{r}=\frac{1}{4\pi {\epsilon}_{0}}\frac{{Q}_{2}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow {Q}_{1}=\frac{r}{R}{Q}_{2}$

The total charge will be

${Q}_{1}+{Q}_{2}=Q\Rightarrow {Q}_{2}\left\{\frac{r}{R}+1\right\}=Q\Rightarrow {Q}_{2}=\left(\frac{R}{R+r}\right)Q\phantom{\rule{0ex}{0ex}}\Rightarrow {Q}_{1}=\left(\frac{r}{R+r}\right)Q$

Therefore, when the two charges are placed at a distance *"d"*, then the force of repulsion between them is,

$F=\frac{1}{4\pi {\epsilon}_{0}}\frac{{Q}_{1}{Q}_{2}}{{d}^{2}}=\frac{1}{4\pi {\epsilon}_{0}}\frac{1}{{d}^{2}}\left(\frac{r}{r+R}\right)\left(\frac{R}{r+R}\right){Q}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow F=\frac{1}{4\pi {\epsilon}_{0}}\frac{{Q}^{2}}{{d}^{2}}\left\{\frac{rR}{{\left(r+R\right)}^{2}}\right\}\phantom{\rule{0ex}{0ex}}ForFtobemaximum\frac{dF}{dR}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dF}{dR}=\frac{d}{dR}\left\{\frac{1}{4\pi {\epsilon}_{0}}\frac{{Q}^{2}}{{d}^{2}}\left\{\frac{rR}{{\left(r+R\right)}^{2}}\right\}\right\}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\left(r+R\right)}^{2}r+rR2\left(r+R\right)}{{\left(r+R\right)}^{4}}=0\Rightarrow {\left(r+R\right)}^{2}r+rR2\left(r+R\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(r+R\right)+2R=0\Rightarrow 3R=r\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{r}{3}$

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