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Two small -sized identical equally charged spheres each having mass 1 mg are hanging in equilibrium as shown in the figure . The length of each string is 10 cm and the angle theta is 7^(@) with the vertical . Calculate the magnitude of the charge in each sphere . (Take g =10ms^(-2)) |
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Answer» Solution :If the spheres are neutral the angle between them will be `0^(@)` when HANGED vertically . SINCE they are positively charged spheres , there will be a repulsive force between them and they will be at equilibrium with each other at an angle of `7^(@)` with the vertiction. We can draw a free body diagram for one of the charged spheres and apply Newton.s second law for both vertical and horizontal directions . The free body diagram is shown. In the x -direction the ACCELERATION of the charged sphere is zero . Using Newton.s second law `(vecF_(m)=mveca)` , we have T sin `thetahati-F_(e)hati=0` T`sin THETA=F_(e)` Here T is the tension acting on the charge due to the string and `F_(e)` is the electrostatic force between the two charges . In the y -direction also the net acceleration experienced by the charge is zero . T `cos thetahatj-mg hatj=0` Therefore T cos `theta =mg ` . By DIVIDING equation (1) by equation (2), `tan theta = (F_(e))/(mg)` Since they are equally charged the magnitude of the electrostatic force is `F_(e)= k""(q_(2))/(r^(2))` where k `=(1)/(4piepsilon_(0))` Here r = 2a = 2L `sin theta`. By substituting these values in equation (3) , `tan theta = k(q^(2))/(mg(2L sin theta)^(2))` Rearranging the equation (4) to get q `q=2L sinthetasqrt(("mg tan" theta)/(k))=2xx0.1xxsin7^(@)xxsqrt(10^(-3)xx10xxtan7^(@))/(9xx10^(9))` `q = 8.9xx10^(-9)C = 8.9nC` 
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