91Ó°ÊÓ

The given data can be listed below,

• The radius of charged disks is,$R$
• The charge on left disk is,$-Q_{left}$
• The charge on the right disk is,$Q_{right}$
• The charge on the rod is $Q_{rod}$
• The length of the rod is,L

The electrostatic force is the attraction (in the case of unlike charges) or repulsion (in the case of like charges) between two-point charges separated by a distance in a vacuum or any dielectric medium at rest.

The electric field on point x when rod lies in x-direction is given by,

$$\begin{gathered} E_x=E_{left}+E_{right} \\ =\frac{Q_{right}}{2A{\mathrm{Î\mu }}_0}-\frac{Q_{left}}{2A{\mathrm{Î\mu }}_0} \end{gathered}$$ …(¾±)

Here,$Q_{right},Q_{left}$, are the charges on right and left disk respectively, A is the area of plate and${\mathrm{Î\mu }}_0$is the permittivity of free space.

The electric field of rod in the y-direction from point x is given by,

$E=\frac{K{Q}_{rod}}{R/2\sqrt{{\left(R/2\right)}^2+{\left(L/2\right)}^2}}$ …(¾±¾±)

Here,$R/2$ is the distance of point x from rod, K is the coulomb constant,$Q_{rod}$ is the charge on the rod and L is the length of the rod.

The magnitude of electric field on the rod is given by,

$\left|E\right|=\sqrt{E_x^2+E_y^2}$

Substitute values from equation (i) and (ii).

$$\begin{gathered} E=\sqrt{{\left(\frac{Q_{right}}{2A{\mathrm{Î\mu }}_0}+\frac{Q_{left}}{2A{\mathrm{Î\mu }}_0}\right)}^2+{\left(\frac{K{Q}_{rod}}{R/2\sqrt{{\left(R/2\right)}^2+{\left(L/2\right)}^2}}\right)}^2} \\ =\sqrt{{\left(\frac{Q_{right}}{2A{\mathrm{Î\mu }}_0}+\frac{Q_{left}}{2A{\mathrm{Î\mu }}_0}\right)}^2+{\left(\frac{K{Q}_{rod}}{R/2\sqrt{{\left(R/2\right)}^2+{\left(L/2\right)}^2}}\right)}^2} \end{gathered}$$

Thus, the magnitude of the electric field of the rod is and it is upward and left pointing.

$\sqrt{{\left(\frac{Q_{right}}{2A{\mathrm{Î\mu }}_0}+\frac{Q_{left}}{2A{\mathrm{Î\mu }}_0}\right)}^2+{\left(\frac{K{Q}_{rod}}{R/2\sqrt{{\left(R/2\right)}^2+{\left(L/2\right)}^2}}\right)}^2}$

The force on the electron at point x is given by,

$F=eE$

Here, e is the charge on the electron and E is the electric field at point x.

Substitute values in the above,

$F=e\sqrt{{\left(\frac{Q_{right}+Q_{left}}{2A{\mathrm{Î\mu }}_0}\right)}^2+{\left(\frac{K{Q}_{rod}}{R/2\sqrt{{\left(R/2\right)}^2+{\left(L/2\right)}^2}}\right)}^2}$

Thus, the electric force exerted on the electron placed between the gap is

$e\sqrt{{\left(\frac{Q_{right}+Q_{left}}{2A{\mathrm{Î\mu }}_0}\right)}^2+{\left(\frac{K{Q}_{rod}}{R/2\sqrt{{\left(R/2\right)}^2+{\left(L/2\right)}^2}}\right)}^2}$