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Chapter 20: Problem 36

A 2 -g ping-pong ball rubbed against a wool jacket acquires a net positive charge of \(1 \mu \mathrm{C}\). Estimate the fraction of the ball's electrons that have been removed.

Short Answer

Expert verified
The fraction of electrons lost by the ping-pong ball during the rubbing process could be found by calculating the number of lost electrons and dividing that by the total number of electrons in a neutral ping-pong ball.

Step by step solution

01

Calculate charge in Coulombs

Using the given data, we know that the ball acquires a net positive charge of \(1 \mu \mathrm{C}\). We can convert \(1 \mu \mathrm{C}\) to Coulombs (C) as follows: \(1 \mu \mathrm{C} = 1 \times 10^{-6} C\). So, the ball's lost electrons have a total charge of \(1 \times 10^{-6} C\).
02

Calculate the number of lost electrons

The loss of electrons is what has provided this charge. To find out the number of lost electrons, we divide the total charge by the charge of one electron. So, \(\text{Total lost electrons} = \frac{(1 \times 10^{-6} C)}{(1.6 \times 10^{-19} C)}\).
03

Calculating the total electrons in the ball

A neutral ball would have just as many electrons as the number of protons otherwise it will have some net charge. The volume of a ping-pong ball is approximately \(33.51 \, cm^3\) and the density of the plastic material for a ping-pong ball is approximately \(0.84 \, g/cm^3\). Giving it a mass of \(33.51 \times 0.84 = 28.1484 \, g\). In each gram of plastic, the number of atoms are approximately \(6.02 \times 10^{23}\). As each atom contributes one electron in a neutral body, this gives us the total electrons in the ball to be \(28.1484 \times 6.02 \times 10^{23}\).
04

Calculate the fraction of lost electrons

We can now calculate the fraction of lost electrons. The number of lost electrons is divided by the originally total electrons in the ping-pong ball.

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Most popular questions from this chapter

A proton is at the origin and an ion is at \(x=5.0 \mathrm{nm}\). If the electric field is zero at \(x=-5\) nm, what's the ion's charge?

The water molecule's dipole moment is \(6.2 \times 10^{-18} \mathrm{C} \cdot \mathrm{m}\). What would be the separation distance if the molecule consisted of charges \(\pm e\) ? (The effective charge is actually less because \(\mathrm{H}\) and O atoms share the electrons.)

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Show that the electric field at a point \(45^{\circ}\) from a dipole's axis is \(1 / \sqrt{5 / 8}\) times the field on the axis, assuming both points are the same distance from the dipole and that distance is large compared with the dipole spacing.

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