InterviewSolution
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InterviewSolution
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A particle A havinga charge of 5.0 xx 10^(-7)C is fixed in a vertical wall. A second particle B of mass 100g and having equal charge is suspended by a silk thread of length 30 cm from the wall. The point of suspension is located 30 cm vertically above the first particle. Find the angle of the thread with the vertical when it stays in equilibrium. |
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Answer» Solution :The situation is show in the figure. Let the point of suspension be O, where the thread makes an angle `theta` with the vetical. FORCES on the particle B are (1) weight mg acting downward. (2) tension, T alongthe thread, (3) electric force of repulsion F along AB. At equilibrium sum of all these forces becomes ZERO. From the given figure, OA=OB, `angleOAB = angleOAB =(90^(@)-theta/2)` Considering the components along BX, we get `F cos theta/2 = mg cos (90^(@)-theta)= mg sin theta= 2mg sin theta/2 cos theta/2` or `sin theta/2 = F/(2mg)` Now, `F=1/(4pi epsilon_(0)).(q_(1)q_(2))/(AB^(2))` and AB = 2(OA) `sin theta/2` `therefore sin theta/2 = 1/(4pi epsilon_(0)).(q_(1)q_(2))/(4(OA)^(2) sin^(2) theta/2). 1/(2mg)`  or `sin^(3) theta/2 =1/(4pi epsilon_(0)).(q_(1)q_(2))/(4(OA)^(2) sin^(2)theta/2). 1/(2mg)` `=(9 XX 10^(9)) xx ((5 xx 10^(-7))^(2)/(4 xx (30 xx 10^(-2))^(2))) xx 1/(2 xx (100 xx 10^(-3)) xx 9.8)` =0.0032 or, `sin theta/2 = 0.15` `therefore theta = 17^(@)` |
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