Question: In Fig. 22-59, an electron (e) is to be released from rest on the central axisof a uniformly charged disk of radiusR. The surface charge density on the disk is$+4.00mC/m^2$. What is the magnitude of the electron’s initial acceleration if it is released at a distance (a)R, (b) R/100, and (c) R /1000from the center of the disk? (d) Why does the acceleration magnitude increase only slightly as the release point is moved closer to the disk?

• a)An electron is released from rest on the central axis of uniformly charged disk of radius, R.
• b) The surface charged density,$\sigma =4mC/m^2$

Using the concept of the electric field and the given data of surface density, we can get the electric field at the given distances can be found.

Formulae:

The magnitude of the electric field produced by the disk at a point on its central axis, $E=\frac{\sigma }{2{\epsilon }_0}\left(1-\frac{Z}{\sqrt{Z^2+R^2}}\right)\left(i\right)$

where,= surface charge density

z = distanceon the central axis of the disk

R = Radius of the disk

The force due to Newton’s second law, F= ma (ii)

The force relation to the electric field,F=qE (iii)

The magnitude of the acceleration at a distance R/100 using equations (i) and (iii) in equation (i) can be given as:

$$\begin{gathered} a=\frac{e\sigma \left(2-\sqrt{2}\right)}{4m{\epsilon }_0} \\ =1.16\times {10}^{16}m/s^2 \end{gathered}$$

Hence, the value of the acceleration is$1.16\times {10}^{16}m/s^2$

The magnitude of the acceleration at a distance R/100 using equations (i) and (iii) in equation (i) can be given as:

$$\begin{gathered} a=\frac{e\sigma \left(10001-\sqrt{10001}\right)}{20002m{\epsilon }_0} \\ =3.94\times {10}^{16}m/s^2 \end{gathered}$$

Hence, the value of the acceleration is$3.94\times {10}^{16}m/s^2$

The magnitude of the acceleration at a distance R/1000 using equations (i) and (iii) in equation (i) can be given as:

$$\begin{gathered} a=\frac{a\sigma \left(1000001-\sqrt{1000001}\right)}{200000m{\epsilon }_0} \\ =3.97\times {10}^{16}m/s^2 \end{gathered}$$

Hence, the value of the acceleration is$3.97\times {10}^{16}m/s^2$

The field due to the disk becomes more uniform as the electron nears the center point. One way to view this is to consider the forces exerted on the electron by the charges near the edge of the disk; the net force on the electron caused by those charges will decrease due to the fact that their contributions come closer to cancelling out as the electron approaches the middle of the disk.