91Ó°ÊÓ

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,$\mathrm{z}=0.8\mathrm{mm}\left(0.8×{10}^{-3}\mathrm{m}\right)$
• The force experienced by glass sheet is,$\mathrm{F}=5×{10}^{-5}\mathrm{N}/\mathrm{C}$

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.

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.

The approximate electric field on the surface of disk,

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

Rearrange the above equation,

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

Substitute all the values in the above,

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

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