Question

A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found that a charged rod can be held above the balloon at a distance of d = 0.0640 m to make the balloon float.
1) In order for this to occur, what polarity of charge must the rod possess?
q2 is ___________

a) negative
b) proton
c) neutral
d) positive

2) How much charge, q2, does the rod have?

Assume the balloon and rod to be point charges. The Coulomb force constant is 1/(4o)8.99 x10 N m2/C2 and the acceleration due to gravity g-9.81 m/s2 O negative 42 is proton O neutral positive Number Tools x 102

Answer 1

Answer:

1)  $\rm q_2$ is positive.

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

Part 1:

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is positive.

Part 2:

Given:

• Mass of the balloon, m = 0.00275 kg.
• Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
• Distance between the rod and the balloon, d = 0.0640 m.
• Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$