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Two indentically charged spheres are suspended by strings of equal lengths. The strings make an angle of 30^(@) with each other. When suspended in a liquid of density 0.8 g. cm^(-3), the angle remains the same. What is the dielelectric constant of the liquid. The density of the material of the spheres is 1.6 g. cm^(-3) |
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Answer» Solution :LET A and B be the equilibrium positions of the two spheres in both cases [Fig.] Suppose, AB=r, charge of each sphere =q, mass of each sphere =m. When thesystem is in air, tension in the string `=T_(1)` and repulsive force between the spheres `F_(1) =1/(4PI epsilon_(0))q^(2)/r^(2)` At equilibrium, `F_(1) =T_(1) sin 15^(@)`..........(1) and `mg = T_(1) cos 15^(@)`........(2) From EQUATIONS (1) and (2), `F_(1)/(mg) =(T_(1)sin 15^(@))/(T_(1) cos 15^(@)) = tan 15^(@)` or `F_(1) = mg tan15^(@)` or `1/(4pi epsilon_(0)).q^(2)/r^(2) mg tan 15^(@)`  When the spheres are suspended in the liquid, tension in the string `=T_(2)`, UPTHRUST by the liquid on each sphere =u and repulsive force between the two spheres, `F_(2) =1/(4pi epsilon_(0)).q^(2)/r^(2)` [where k=dielectric constant of the medium]. Now at equilibrium, `F_(2) = T_(2) sin 15^(@)`......(4) and `mg-u = T_(2) cos 15^(@)`.........(5) From equations (4) and (5), `F_(2)/(mg -u) = tan 15^(@)` or, `F_(2) =(mg-u) tan 15^(@)` or, `1/(4pi epsilon_(0)k).q^(2)/r^(2) =(mg-u) tan 15^(@)`.........(6) Form equations (3) and (6), we get `k = (mg)/(mg-u)`....(7) Let volume of the sphere =V, the density of the material of the sphere =d and density of the liquid `=d_(l)`. `therefore m = Vd` and `u=Vd_(1)g)` HENCE, from equations (7) , we get `x=(Vdg) /(Vdeg-Vd_(l)g) =d/(d-d_(1))` or, `x =(1.6)/(1.6 -0.8)[therefore d=1.6 g//cm^(3)` and `d_(l) = 0.8 g//cm^(3)]` or, k=2 Hence, required dielectric constant is 2. |
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