Question

You hold a ruler that has a charge on its tip 3.60 cm above a small piece of tissue paper to see if it can be picked up. The ruler has −16.0 µC of charge. The tissue paper has 5.00 g of mass. What is the minimum charge required to pick up the tissue paper?

Answer 1

The minimum charge required to pick up the tissue paper is 44.1 nC.

Step-by-step explanation:

1-The force of attraction between the ruler and tissue paper is given by Coulomb's Law of Electrostatics which is as

`F=(k q_1 q_2)/(r^2)`

Here

• F is the force required to pick the tissue paper up. The minimum force required is equal to the weight of tissue paper. which is given in SI as

`F= mg\\F=(5.00)/(1000) * 9.8 N \\ F=0.049 N`

• k is the constant value which is given as

`k=9.0 * 10^9 (N m^2)/(C)`

• 
  `q_1` is the charge on the ruler which is -16μC.
• 
  `q_2` is the charge on the tissue paper which is to be calculated.
• r is the distance between two bodies which is given as 3.60 cm.

Putting values and solving in the equation gives

`F=(k q_1 q_2)/(r^2) \\ -0.049 N=(9.0 * 10^9 -16 * 10^(-6) * q_2)/(((3.60)/(100))^2) \\ -0.049 N=(-1.44* 10^(9-6+2) * q_2)/(1.296 * 10^(-3)) \\ -0.049 N={-1.11* 10^(3+2+3) * q_2} \\ -0.049 N={-1.11* 10^(8) * q_2} \\ q_2= ( -0.049 )/(-1.11* 10^(8)) \\ q_2=4.41 * 10^(-10) C \\ q_2=44.1 * 10^(-9) C \\q_2=44.1 nC`