Question

Inez is putting up decorations for her sister's quinceanera (fifteenth birthday party). She ties three light silk ribbons together to the top of a gateway and hangs from each ribbon a rubber balloon. To include the effects of the gravitational and buoyant forces on it, each balloon can be modeled as a particle of mass m=2.79 g, with its center 48.7 cm from the point of support. To show off the colors of the balloons, Inez rubs the whole surface of each balloon with her woolen scarf, to make them hang separately with gaps between them. The centers of the hanging balloons form a horizontal equilateral triangle with sides 30.2 cm long. What is the common charge each balloon carries?

Answer 1

Let Fe be the electrostatic force between one balloon and another. That force will be horizontal. Looking down from above, The horizontal force on any one balloon is the vector sum of the forces from the adjacent balloons. An equilateral triangle's angle is 60º, so the angle between the forces is 60º. The resultant force will bisect that angle, and the magnitude will be 2*Fe*cos30º. That force must be balanced by the horizontal component of the weight of the balloon, W = m*g. That component will depend on the angle the ribbon makes with the horizontal. The tie point of the ribbon is directly above the geometric center of the triangle. The distance from a vertex to the center of an equilateral triangle is L/(2*cos30º) if L is the side of the triangle. The ribbon is the hypotenuse and the vertex distance the adjacent side to the angle the ribbon makes with the horizontal. If the ribbon length is R, then that angle is ø = arccos(L/(2*cos30º) / R) ø = arccos L/(2*cos30º*R).
For L = 30.2 and R = 48.7,
ø = arccos L/(2*cos30º*R)
= arccos(30.2/(2*cos30*48.7))
= 69.02 deg

The vertical component of tension in the ribbon must support the vertical weight of the balloon, so Tr *sinø= m*g; Tr = m*g/sinø. The horizontal component of Tr is Tr*cosø, so Fh = m*g*cotø

Now equate Fh with the resultant electrostatic force

m*g*cotø = 2*Fe*cos30º

0.00279*9.8*0.3835 = 2*Fe*0.866

Fe = 0.00605 N

To produce that force between two charges of value q:

Fe = K*q²/L² (L is the distance between the charges, the length of the side of the triangle).

q = √[Fe*L²/K]

q = L*√[Fe/K] where K = 9*10^9 N*m²/C²

q = 0.05*√[0.00605/9*10^9]
= 4.1*10^-8 C