Question

When a charged balloon sticks to a wall, the downward gravitational force is balanced by an upward static friction force. The normal force is provided by the electrical attraction between the charged balloon and the equal but oppositely charged polarization induced in the wall's molecules.If the mass of a balloon is 1.9 g, its coefficient of static friction with the wall is 0.70, and the average distance between the opposite charges is 0.55 mm, what minimum amount of charge must be placed on the balloon in order for it to stick to the wall?

Answer 1

Answer:

$9.465534174\times 10^{-10}\ C$

Explanation:

m = Mass of balloon = 1.9 g

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

g = Acceleration due to gravity = 9.81 m/s²

r = Distance = 0.55 mm

$\mu$ = Coefficient of friction = 0.7

q = Charge

The gravitational force will balance the electrical force

$mg=\dfrac{kq^2}{r^2}\\\Rightarrow m\dfrac{g}{\mu}=\dfrac{kq^2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{mgr^2}{k\mu}}\\\Rightarrow q=\sqrt{\dfrac{1.9\times 10^{-3}\times 9.81\times (0.55\times 10^{-3})^2}{8.99\times 10^9\times 0.7}}\\\Rightarrow q=9.465534174\times 10^{-10}\ C$

The charge is $9.465534174\times 10^{-10}\ C$