Question

Two identical helium filled balloons A and B fastened to a weight of 5 g by threads floats in equilibrium as shown in figure. Calculate the charge on each balloons,assuming that they carry equal charges.
Ans : q = ± 5.6 ×
${10}^{ - 7}$
​

Answer 1

$\large\mathbb{\underbrace{\red{SOLUTION:-}}}$

☞ See the attachment figure .

GIVEN :-

✍️ Weight of the hanging mass is ‘5g’ or ‘$\rm\bold{5\times{10^{-3}}\:kg}$’ .

✍️ In the attachment figure, A and B are two balloons attached to weight ‘W’ by strings ‘OA’ and ‘OB’ .

• AB = 0.6m

• OA = OB = 1m

☞ So,

• AC = BC = AB/2 = 0.6/2 = 0.3m

☞ $\rm{\therefore\:\:OC\:=\:\sqrt{(OA)^2\:-\:(AC)^2}}$

$\rm{\implies\:OC\:=\:\sqrt{1^2\:-\:(0.3)^2\:}\:=\:0.954\:}$

CALCULATION :-

✍️ Now using ∆ law of vectors, we get

$\checkmark\:\rm{\purple{\dfrac{F}{AC}\:=\:\dfrac{w}{OC}\:}}$ ------(a)

☞ where,

• F is the repulsive electrostatic force acting on the charged balloons .

• w is the up-thrust experienced by balloons due to Buoyancy .

☞ As, [T cosθ = w] -------(1)

☞ Also, 2T cosθ = W => T cosθ = W/2 .

☞ Hence, from equation (1) ‘w = W/2’ .

[NOTE :- W = mg ]

✍️ Now put these values in the equation (a), we get

$\rm{\implies\:\dfrac{F}{AC}\:=\:\dfrac{W/2}{OC}\:}$

$\rm{\implies\:F\:=\:\dfrac{AC}{OC}\times{\dfrac{W}{2}}\:}$

$\rm{\implies\:\dfrac{q^2}{4\pi\epsilon_{o}\:(AB)^2}\:=\:\dfrac{AC}{OC}\times{\dfrac{mg}{2}}\:}$

$\rm{\implies\:q^2\:=\:4\pi\epsilon_{o}\times{(AB)^2}\times{\dfrac{AC}{OC}}\times{\dfrac{mg}{2}}\:}$

$\rm{\implies\:q^2\:=\:\dfrac{(0.6)^2\times(0.3)\times(5\times{10^{-3}}\times{9.8})}{(9\times{10^9})\times{0.954}\times{2}}\:}$

$\rm{\implies\:q^2\:=\:\dfrac{0.36\times{0.3}\times{49}}{9\times{0.954}\times{2}}\:\times{10^{-12}}}$

$\rm{\implies\:q\:=\:±0.56\times{10^{-6}}\:C}$

$\bigstar\:\rm{\boxed{\blue{\implies\:q\:=\:±5.6\times{10^{-7}}\:C}}}$

Answer 2

$\large\mathbb{\underbrace{\red{SOLUTION:-}}}$

☞ See the attachment figure .

GIVEN :-

✍️ Weight of the hanging mass is ‘5g’ or ‘\rm\bold{5\times{10^{-3}}\:kg}[/tex]

✍️ In the attachment figure, A and B are two balloons attached to weight ‘W’ by strings ‘OA’ and ‘OB’ .

AB = 0.6m

OA = OB = 1m

☞ So,

AC = BC = AB/2 = 0.6/2 = 0.3m

☞ $\rm{\therefore\:\:OC\:=\:\sqrt{(OA)^2\:-\:(AC)^2}}$

$\rm{\implies\:OC\:=\:\sqrt{1^2\:-\:(0.3)^2\:}\:=\:0.954\:}$

CALCULATION :-

✍️ Now using ∆ law of vectors, we get

$\checkmark\:\rm{\purple{\dfrac{F}{AC}\:=\:\dfrac{w}{OC}\:}}$

☞ where,

F is the repulsive electrostatic force acting on the charged balloons .

w is the up-thrust experienced by balloons due to Buoyancy .

☞ As, [T cosθ = w] -------(1)

☞ Also, 2T cosθ = W => T cosθ = W/2 .

☞ Hence, from equation (1) ‘w = W/2’ .

[NOTE :- W = mg ]

✍️ Now put these values in the equation (a), we get

$\rm{\implies\:\dfrac{F}{AC}\:=\:\dfrac{W/2}{OC}\:}$

$\rm{\implies\:F\:=\:\dfrac{AC}{OC}\times{\dfrac{W}{2}}\:}$

$\rm{\implies\:\dfrac{q^2}{4\pi\epsilon_{o}\:(AB)^2}\:=\:\dfrac{AC}{OC}\times{\dfrac{mg}{2}}\:}$

$\rm{\implies\:q^2\:=\:4\pi\epsilon_{o}$$\times{(AB)^2}\times{\dfrac{AC}{OC}}$$\times{\dfrac{mg}{2}}\:}[tex]</p><p> </p><p></p><p>[tex]\rm{\implies\:q^2\:=\:$$\dfrac{(0.6)^2\times(0.3)\times(5\times{10^{-3}}$$</p><p>[tex]\times{9.8})}{(9\times{10^9})\times{0.954}\times{2}}\:}$

$\rm{\implies\:q^2\:=\:\dfrac{0.36\times{0.3}\times{49}}{9\times{0.954}\times{2}}\:\times{10^{-12}}}$

$\rm{\implies\:q\:=\:±0.56\times{10^{-6}}\:C}$

[tex]\bigstar\:\rm{\boxed{\blue{\implies\:q\:=\:±5.6\times{10^{-7}}\:C}}}★[tex]