Question

In Fig. 23-48a, an electron is shot directly away from a uniformly charged plastic sheet, at speed ${\mathit{V}}_{\mathbf{5}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathit{m}\mathbf{/}\mathit{s}$ . The sheet is non-conducting, flat, and very large. Figure 23-48bgives the electron’s vertical velocity component vversus time tuntil the return to the launch point. What is the sheet’s surface charge density?

Short answer

The surface charge density of the sheet is $2.9\times {10}^{-6}C/m^2$.

Step-by-step solution

The given data

1. An electron is shot directly away at speed,$v_s=2.0\times {10}^{5}m/s$
2. The sheet is non-conducting, flat, and very large.

Understanding the concept of the electric field and Newton’s law

Using the concept of the electric field of a non-conducting sheet with the electrostatic force value of Newton's second law, we can get the expression for acceleration. Using this equation, we can get the surface charge density of the sheet. Again, we know that the slope of velocity and time graph gives the acceleration, using this value, charge, and mass of the electron, we get the required answer.

Formulae:

The electric field of a non-conducting sheet, $E=\frac{\sigma }{2{\epsilon }_0}$ (1)

The force due to Newton’s second law, $F=ma$ (2)

The electrostatic force due to passing electric field,$F=qE$(3)

Calculation of the surface charge density of the sheet

Substituting the value of the electric field from equation (1) in equation (3) and then substituting this force equation in equation (2), we get the acceleration value as given:

$$\begin{gathered} a=\frac{e\sigma }{2{\epsilon }_{0}m} \\ \frac{2.0\times {10}^{5}m/s}{7.0\times {10}^{-12}s}=\frac{(1.6\times {10}^{-19}C)}{2\times (8.85\times {10}^{-12}F/M)\times (9.1\times {10}^{-31}KG} \\ \sigma =2.9\times {10}^{-6}C/m^2 \end{gathered}$$

Hence, the value of the surface charge density of the sheet is $2.9\times {10}^{-6}C/m^2$ .