Question

Question: A charged nonconducting rod, with a length of 2.00 mand a cross-sectional area of$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{2}}$, lies along the positive side of ax-axis with one end at the origin. The volume charge density is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if $\mathit{\rho }$ is (a) uniform, with a value of$\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathit{\mu }\mathbf{}\mathit{C}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$, and (b) non-uniform, with a value given by$\mathit{r}\mathbf{}\mathbf{=}\mathit{b}{\mathit{x}}^{\mathbf{2}}$, where$\mathit{b}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathit{\mu }\mathbf{}\mathit{C}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$?

Short answer

Answer:

1. The number of excess electrons on the rod if r is uniform is$2.00\times {10}^{10}$.
2. The number of excess electrons on the rod if r is non-uniform is$1.33\times {10}^{10}$.

Step-by-step solution

Stating given data 

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

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

Understanding concept of quantization of charge

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,$dq=\rho AL\text{or}\rho Adx$.(i)

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

a) Calculation of excess electrons when density is uniform

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

$$\begin{gathered} n=\frac{\left|\rho \right|A}{e}{\int }_o^{L}dx \\ d=\frac{\left|\rho \right|AL}{e} \\ =2.00\times {10}^{10} \end{gathered}$$

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

b) Calculation of excess electrons when density is non-uniform

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

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

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