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Two identical helium filled ballons  A and B fastened to a weight of 5g by threads floats in equilibrium as shown in figure.Calculate the charge on each ballons,assuming that they carry equal charges.please explain .Ans is 5×10-7C

Asked by valavanvino1011 25th April 2019, 12:33 AM
Answered by Expert
Figure shows the forces acting on hanging weight and He Balloons. 
Fe is the repulsive electrostatic force acting onthe charged baloons .
Ft is the upthrust experienced by baloons due to bouyancy. Tension forces acting on the strings is also shown in figure.
It can be seen from figure, net vertical component of Tension is balanced by weight of the hanging mass.
Hence we have, 2 T cosθ = m g    or   T =  mg / (2 cosθ)  ...................(1)
at He balloon, horizontal component of Tension is balanced by electrostatic force, 
Hence we have,   Fe = T sinθ .................(2)
Using eqn.(1), Eqn.(2) is written as,  Fe = begin mathsize 12px style fraction numerator q squared over denominator 4 πε subscript 0 straight r squared end fraction space equals space fraction numerator m g over denominator 2 end fraction tan theta space space space space space o r space space space space space q space equals square root of m space g space 2 space straight pi space straight epsilon subscript 0 space straight r squared space tanθ end root end style .........................(3)
where q is the charge on each balloon, m is mass of hanging object, r is the distance between balloon,
g is acceleration due to gravity, ε0 is the permitivity of free space.
we can see from figure, tanθ = begin mathsize 12px style fraction numerator 0.3 over denominator square root of 1 minus open parentheses 0.3 close parentheses squared end root end fraction space almost equal to space 0.31 end style
By substituting all the values in eqn.(3), we get q = 5.5×10-7 C
(distance between balloons is assumed as 0.6 m, not 0.6 cm as shown in figure.  
0.6 cm or 6 mm distance between balloons, precisely between centres of balloons,  is negligibly small distance , practically not possible )
Answered by Expert 25th April 2019, 10:31 AM
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