Question

A thin circular sheet of glass of diameter 3 m is rubbed with a cloth on one surface and becomes charged uniformly. A chloride ion (a chlorine atom that has gained one extra electron) passes near the glass sheet. When the chloride ion is near the center of the sheet, at a location 0.8 mm from the sheet, it experiences an electric force of 5 × 10−15 N, toward the glass sheet. It will be useful to you to draw a diagram on paper, showing field vectors, force vectors, and charges, before answering the following questions about this situation. Which of the following statements about this situation are correct? Select all that apply. (1) The electric field that acts on the chloride ion is due to the charge on the glass sheet and to the charge on the chloride ion. (2) The electric field of the glass sheet is equal to the electric field of the chloride ion. (3) The charged disk is the source of the electric field that causes the force on the chloride ion. (4) The net electric field at the location of the chloride ion is zero. (5) The force on the chloride ion is equal to the electric field of the glass sheet. In addition to an exact equation for the electric field of a disk, the text derives two approximate equations. In the current situation we want an answer that is correct to three significant figures. Which of the following is correct? We should not use an approximation if we have enough information to do an exact calculation. (1) R≫z, so it is adequate to use the most approximate equation here. (2) z is nearly equal to R, so we have to use the exact equation. (3) z≪R, so we can’t use an approximation. How much charge is on the surface of the glass disk? Give the amount, including sign and correct units

Short answer

Option (3) is correct and the charge on the surface of glass disk is$3.9\times {10}^{-6}\mathrm{C}$ .

Step-by-step solution

Identification of given data

The given data can be listed below,

• The diameter of circular glass disk is,R=3m
• The radial distance of point from center of disk is,$\mathrm{z}=0.8\mathrm{mm}\left(0.8\times {10}^{-3}\mathrm{m}\right)$
• The force experienced by glass sheet is,$\mathrm{F}=5\times {10}^{-5}\mathrm{N}/\mathrm{C}$

Concept/Significance of electrostatic force

When an electric charge in a certain reference frame is stationary. Then the term "electrostatic field" or "force" is appropriate. Both electrostatic and magnetic fields and forces are present when a charge is moving.

Determination of the true statement.

The claims in the problem:

1. The electric field that acts on the ion is caused only by the charge on the disk, so this statement is false. The force is caused by both charged because if the ion were neutral, it would not feel a force.
2. Due to Newton's third law, the force exerted on the ion by the charged disk is the same as the force exerted on the disk by the ion. Since their charges are different (we calculate this later), the electric fields have to be different, so this statement is false.
3. This statement is true as the chloride ion has a charge and since all other charges cancel, the chloride ion effectively acts like just one electron in the presence of an external field. The ion feels a force because of the charged disk and the force is

Here, e is the charge on the ion and E is the electric field.

1. Since the force on the ion is non-zero, the field at the position of the ion cannot be zero, so this statement is false.
2. The force on the ion is equal to the electric field times the charge of the ion, so this statement is false.

Step 4: Determination of charge is on the surface of the glass disk

The approximate electric field on the surface of disk,

$E=\frac{\left(Q/A\right)}{2{\epsilon }_0}$

Rearrange the above equation,

$$\begin{gathered} F=eE \\ =\frac{e\left(Q/A\right)}{2{\epsilon }_0} \\ Q=\frac{2\epsilon AF}{e} \end{gathered}$$

Substitute all the values in the above,

$$\begin{gathered} \mathrm{Q}=\frac{2\left(8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right){\left(1.5\mathrm{m}\right)}^2\mathrm{\pi }\left(5\times {10}^{-15}\mathrm{N}\right)}{1.6\times {10}^{-19}\mathrm{C}} \\ =3.9\times {10}^{-6}\mathrm{C} \end{gathered}$$

Thus, the force is attractive the charge on ion negative so the charge on disk must be positive with value$3.9\times {10}^{-6}\mathrm{C}$ .