Question

Two light gas balloons, each with charge +q are tied to a wooden stick of
mass 2 grams using an identical massless string of length 1 m. Calculate the
charge on both the balloons, if distance between both the balloons floating in
air is 40 cm at equilibrium.​

Answer 1

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.

. Hence we have, 2 T cosθ = m g or T = mg / (2 cosθ) ...................(1)

. Hence we have, 2 T cosθ = m g or T = mg / (2 cosθ) ...................(1)

. Hence we have, 2 T cosθ = m g or T = mg / (2 cosθ) ...................(1) .

. Hence we have, 2 T cosθ = m g or T = mg / (2 cosθ) ...................(1) .

. Hence we have, 2 T cosθ = m g or T = mg / (2 cosθ) ...................(1) . 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

. Hence we have, 2 T cosθ = m g or T = mg / (2 cosθ) ...................(1) . 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 styleBy substituting all the values in eqn.(3), we get q = 5.5×10-7 C

. Hence we have, 2 T cosθ = m g or T = mg / (2 cosθ) ...................(1) . 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 styleBy substituting all the values in eqn.(3), we get q = 5.5×10-7 C

. Hence we have, 2 T cosθ = m g or T = mg / (2 cosθ) ...................(1) . 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 styleBy 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.

. Hence we have, 2 T cosθ = m g or T = mg / (2 cosθ) ...................(1) . 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 styleBy 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.

. Hence we have, 2 T cosθ = m g or T = mg / (2 cosθ) ...................(1) . 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 styleBy 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 )

. Hence we have, 2 T cosθ = m g or T = mg / (2 cosθ) ...................(1) . 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 styleBy 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

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