Question

Two similar balloons filled with helium gas are tied to "L" m long strings. A body of mass m is tied to another ends of the strings.

The balloons float on air at distance "r". If the amount of charge on the balloons is same then the magnitude of charge on each

balloon will be?

Class-12th/Physics
Chapter- Electrostatics
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Answer 1

▶ Solution :

▪ First, see the attachment for better understanding.

✴ Force equations :

• Force equilibrium (whole system) :

▪ Let buoyant force acts on each ballon = B

▪ Net buoyant force = 2B (upward)

▪ Weight force of body = mg (downward)

☞ 2B = mg ....... (1)

• Force equilibrium (For ballon) :

✴ Along x-axis :

☞ Fe = T sinΦ ........ (2)

▪ Fe = Electrostatic force of repulsion

✴ Along y-axis :

☞ B = T cosΦ ......... (3)

→ Taking (2)/(3), we get,

☞ tanΦ = Fe/B

As per equation (1), B = mg/2

☞ tanΦ = 2kQ^2/r^2mg

☞ Q^2 = mgr^2 tanΦ/2k

☞ Q = √(mgr^2 tanΦ/2k)

Answer 2

Gìven :

• Two similar balloons filled with He gas are tied to 'L ' m long strings .
• A body of mass m is tied to another ends of the strings .
• Amount of charge on the ballons is same .
• Balloons float on air at distance 'r'

To Fìnd :

• the magnitude of charge on each balloon.

Solution:

Draw a free body diagram for the whole system. ( Refer to the attachment )

From the Free body Diagram of the system :

Net buoyant force = Weight of the body

⇒ 2B = mg ....(1)

At equlilbrum :

From the free body diagram of balloon :

• Along X - axis

$\sf\:F_{e}=T\sin\theta...(2)$

• Along Y - axis

$\sf\:B=T\cos\theta...(3)$

On diving equation (2) and (3)

$\implies \sf\sf\: \dfrac{F_{e}}{B} = \dfrac{T \sin \theta}{T \cos \theta}$

$\implies \sf\: \dfrac{F_{e}}{B} = \tan \theta$

$\implies \sf \dfrac{kq {}^{2} }{r {}^{2}B} = \tan \theta$

Put the value of B from equation (1)

$\implies \sf \dfrac{2 \times kq {}^{2} }{r {}^{2}mg} = \tan \theta$

$\implies \sf q {}^{2} = \dfrac{mgr {}^{2} \tan \theta}{k \times 2}$

$\implies \sf q = \sqrt{ \dfrac{ \tan \theta \times mgr {}^{2} }{2k} }$

${\red{\boxed{\large{\bold{Note:}}}}}$

Refer to the attachments .