Question

In Figure 15.61 are two uniformly charged disks of radius R that are very close to each other (gap≪R). The disk on the left has a charge of−${\mathit{Q}}_{\mathbf{l}\mathbf{e}\mathbf{f}\mathbf{t}}$and the disk on the right has a charge of +${\mathit{Q}}_{\mathbf{r}\mathbf{i}\mathbf{g}\mathbf{h}\mathbf{t}}$(${\mathit{Q}}_{\mathbf{right}}$is greater than${\mathit{Q}}_{\mathbf{l}\mathbf{e}\mathbf{f}\mathbf{t}}$). A uniformly charged thin rod of length L lies at the edge of the disks, parallel to the axis of the disks and cantered on the gap. The rod has a charge of +${\mathit{Q}}_{\mathbf{r}\mathbf{o}\mathbf{d}}$.

(a) Calculate the magnitude and direction of the electric field at the point marked × at the center of the gap region, and explain briefly, including showing the electric field on a diagram. Your results must not contain any symbols other than the given quantities R,${\mathit{Q}}_{\mathbf{l}\mathbf{e}\mathbf{f}\mathbf{t}}$, ${\mathit{Q}}_{\mathbf{r}\mathbf{i}\mathbf{g}\mathbf{h}\mathbf{t}}$, L, and${\mathit{Q}}_{\mathbf{r}\mathbf{o}\mathbf{d}}$(and fundamental constants), unless you define intermediate results in terms of the given quantities. (b) If an electron is placed at the center of the gap region, what are the magnitude and direction of the electric force that acts on the electron?

Short answer

a) The magnitude of the electric field of the rod is $\sqrt{{\left(\frac{Q_{right}+Q_{left}}{2A{\epsilon }_0}\right)}^2+{\left(\frac{K{Q}_{rod}}{R/2\sqrt{{\left(R/2\right)}^2+{\left(L/2\right)}^2}}\right)}^2}$and its direction is upward and left pointing.

b) Thus, the electric force exerted on the electron placed between the gap is $e\sqrt{{\left(\frac{Q_{right}+Q_{left}}{2A{\epsilon }_0}\right)}^2+{\left(\frac{K{Q}_{rod}}{R/2\sqrt{{\left(R/2\right)}^2+{\left(L/2\right)}^2}}\right)}^2}$.

Step-by-step solution

Identification of given data

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

Concept/Significance of electrostatic force

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.

(a) Determination of the magnitude and direction of the electric field at the point marked x at the center of the gap region

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

$$\begin{gathered} E_x=E_{left}+E_{right} \\ =\frac{Q_{right}}{2A{\epsilon }_0}-\frac{Q_{left}}{2A{\epsilon }_0} \end{gathered}$$ …(i)

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

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

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

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).

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

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

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

(b) the magnitude and direction of the electric force that acts on the electron when it placed between the gap area.

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,

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

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

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