91Ó°ÊÓ

1. Length of the rod,$l=2.00\text{m}$.
2. Cross-sectional area of the rod,$A=4.00{\text{cm}}^{\text{2}}\text{or}4.00×{10}^{-4}{\text{m}}^{\text{2}}$.
3. Uniform density,$\mathrm{ÒÏ}=-4.00\mathrm{μ}C/m^2$.

Non uniform density, $\mathrm{ÒÏ}=b{x}^2$where$b=-2.00\mathrm{μ}C/m^2$.

From the density and area of the substance, we can get the amount of the net charge present in the system. Now, the number of excess electrons can be found by dividing the net charge with the electronic charge value.

Formulae:

Net charge present in the system,$dq=\mathrm{ÒÏ}AL\text{or}\mathrm{ÒÏ}Adx$.(i)

The number of excess electrons,$n=∫\left|dq\right|/e$ .

In the case where$-4.00\mathrm{μ}C/m^2$, the number of excess electrons, on using equation (i) in equation (ii), is found to be

$$\begin{gathered} n=\frac{\left|\mathrm{ÒÏ}\right|A}{e}{∫}_o^{L}dx \\ d=\frac{\left|\mathrm{ÒÏ}\right|AL}{e} \\ =2.00×{10}^{10} \end{gathered}$$

Hence, the number of excess electrons is$=2.00×{10}^{10}$.

The non-uniform density in this case,$\mathrm{ÒÏ}=b{x}^2,\text{where}b=-2.00\mathrm{μ}C/m^2$.So, the number of electrons, using equation (ii) in equation (i) and the given data, can be calculated as follows:

$$\begin{gathered} n=\left|b\right|A{∫}_O^{L}dx \\ =\frac{\left|b\right|A{L}^3}{3e} \\ =1.33×{10}^{10} \end{gathered}$$

Hence, the number of excess electrons is$1.33×{10}^{10}$.