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Three particles, each of mass 1gm and carrying a charge q, are suspended from a common point by insulated massless strings, each 100cm long. If the particles are in equilibrium and are located at the corners of an equilateral triangle of side length 3cm, calculate the charge q on each particle. (Take g=10m//s^2). |
Answer»   Let T make and angle `theta` with the vertical `OC=2/3sqrt((0.03)^2-(0.015)^2)=0.0173m` `:.` `OM=0.9997` NOTE THIS STEP: Resolving T in the direction of mg and F and applying the condition of equilibrium, we get `Tcos theta=mg`, `Tsintheta=F` `:.` `tan theta=(F)/(mg)` ...(i) `F=sqrt(F_(CA)^2+F_(CB)^2+2F_(CA)F_(CB)cosalpha)` `:.` `F=sqrt(F_(CA)^2+F_(CA)^2+2F_(CA)^2xx1/2)` `F=sqrt3F_(CA)=sqrt3xx(kq^2)/((CA)^2)` ...(II) [where `F_(CB)=Force on C due to B F_(CA)=Force on C due to A `|vecF_(CB)|=|vecF_(CA)|` and `alpha=60^@`] ALSO, `tantheta=(OC)/(OM)=(0.0173)/(0.9997)` ...(iii) From (i), (ii) and (iii) `(0.0173)/(0.9997)=(sqrt3xx9xx10^9xxq^2)/((0.03)^2xx10^-3xx9.8)` On SOLVING, we get `q=3.16xx10^-9C`. |
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