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

Answer:

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

Explanation:

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

$F=\frac{k q_1 q_2}{r^2}$

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

                                                     $F= mg\\F=\frac{5.00}{1000} \times 9.8 N \\ F=0.049 N$

• k is the constant value which is given as

                                                 $k=9.0 \times 10^9 \frac{N m^2}{C}$

• $q_1$ 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

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