# Two identical charged spheres are suspended by strings of equal lengths .The string makes an angle of 30 with each other When suspended in liquid of density 0.8g/cm^3, The angle remains the same.if density of material of sphere is 1.6g/cm^3 the dielectric constant of liquid is?

Suppose T is the tension in the string . Force of repulsion  $\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}×\frac{{\mathrm{q}}^{2}}{{\mathrm{r}}^{2}}$
Resolving these forces in the horizontal and vertical components as shown in figure
$\mathrm{Tcos}\left(\mathrm{\theta }\right)=\mathrm{mg}\phantom{\rule{0ex}{0ex}}\mathrm{Tsin}\left(\mathrm{\theta }\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}×\frac{{\mathrm{q}}^{2}}{{\mathrm{r}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{we}\mathrm{get},\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\mathrm{\theta }\right)=\frac{{\mathrm{q}}^{2}}{4{\mathrm{\pi \epsilon }}_{0}{\mathrm{mgr}}^{2}}---\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{immersing}\mathrm{in}\mathrm{the}\mathrm{liquid},\phantom{\rule{0ex}{0ex}}\mathrm{Appearent}\mathrm{weight}\mathrm{mg}\text{'}=\mathrm{mg}\left(1-\frac{0.8}{1.6}\right)=\frac{\mathrm{mg}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Force}\mathrm{of}\mathrm{repulsion}\mathrm{F}\text{'}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}\mathrm{K}}×\frac{{\mathrm{q}}^{2}}{{\mathrm{r}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{If}\mathrm{T}\text{'}\mathrm{is}\mathrm{the}\mathrm{new}\mathrm{tension},\mathrm{then}\phantom{\rule{0ex}{0ex}}\mathrm{T}\text{'}\mathrm{cos}\left(\mathrm{\theta }\right)=\frac{\mathrm{mg}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{T}\text{'}\mathrm{sin}\left(\mathrm{\theta }\right)=\frac{{\mathrm{q}}^{2}}{4{\mathrm{\pi \epsilon }}_{0}{\mathrm{Kr}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}:\mathrm{tan}\left(\mathrm{\theta }\right)=\frac{2{\mathrm{q}}^{2}}{4{\mathrm{\pi \epsilon }}_{0}{\mathrm{Kmgr}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Using}\left(1\right)\phantom{\rule{0ex}{0ex}}\frac{2{\mathrm{q}}^{2}}{4{\mathrm{\pi \epsilon }}_{0}{\mathrm{Kmgr}}^{2}}=\frac{{\mathrm{q}}^{2}}{4{\mathrm{\pi \epsilon }}_{0}{\mathrm{mgr}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{K}=2$

Dielectric constant of liquid is 2

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