Q48P - Fundamentals Of Physics

Question

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

Step-by-Step Solution

Answer

• a) The magnitude of the electron’s initial acceleration at distance, R is$1.16\times {10}^{16} m/s^2$
• b) The magnitude of the electron’s initial acceleration at distance, R/100 is $3.94\times {10}^{16} m/s^2$
• c) The magnitude of the electron’s initial acceleration at distance, R/1000 is $3.97\times {10}^{16} m/s^2$
• d) The contributions from the force increases with decreasing distance, thus, the acceleration magnitude increases.

Step 1: The given data

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

Step 2: Understanding the concept of electric field

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 = distance on 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)

 

Step 3: a) Calculation of the acceleration at R

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$1.16\times {10}^{16} m/s^2$

Step 4: b) Calculation of the acceleration at R/100

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$3.94\times {10}^{16} m/s^2$

Step 5: c) Calculation of the acceleration at R/1000

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 $3.97 \times {10}^{16} m/s^2$

 

Step 6: d) Justifying the acceleration increase with the given cases

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.