The thin plastic rod of length L = 10.0cmin Fig. 24-47 has a non-uniform linear charge density$\mathit{\lambda }\mathbf{=}\mathit{c}\mathit{x}$, where$\mathit{c}\mathbf{=}\mathbf{49}\mathbf{.}\mathbf{9}\mathit{p}\mathit{C}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$. (a) With V= 0 at infinity, find the electric potential at point P₂ on the yaxis at y = D = 3.56cm. (b) Find the electric field component at P₂. (c) Why cannot the field component E_xat P₂ be found using the result of (a)?

1. Length of the thin rod, L = 10cm
2. Non-uniform linear charge density,$\lambda =cx$ where c = 49.9pC/m².
3. Point is present on the y-axis at D = 3.56cm.

Using the electric potential concept of a thin rod, we can get the required electric potential by substituting the given values. Again, this same electric potential relation will determine the electric field of the given potential. This found relation also explains the reason for the electric field by stating the potential which is a function of the y-component.

Formulae:

The potential of a thin rod at the given point on the x-axis, $dV=\frac{1}{4\pi {\epsilon }_0}\frac{dq}{\sqrt{x^2+y^2}}$ (i)

The linear charge density of a body, $\lambda =\frac{q}{L}$ (ii)

The electric field relation to the potential, $E=-\frac{dV}{dx}$ (iii)

This figure shows a thin plastic rod of length L lying on the x-axis.

Considering an infinitesimal segment rod of length dx with a charge of dq, we get the value of the charge using equation (ii) as follows:

$\begin{array}{rcl}dq& =& \lambda dx\\ & =& cxdx\end{array}$

Then, the potential at the given point is obtained using the above data in equation (i) as follows:

$dV=\frac{1}{4\pi {\epsilon }_0}\frac{cxdx}{\sqrt{x^2+y^2}}$

The electric potential at point P₂ is given by integrating the above function as follows:

$\begin{array}{rcl}V& =& \underset{0}{\overset{L}{\int }}dv\\ & =& \frac{1}{4\pi {\epsilon }_0}\underset{0}{\overset{L}{\int }}\frac{cxdx}{\sqrt{x^2+y^2}}\\ & =& \frac{c}{4\pi {\epsilon }_0}\underset{0}{\overset{L}{\int }}\frac{xdx}{\sqrt{x^2+y^2}}\\ & =& \frac{c}{4\pi {\epsilon }_0}\underset{0}{\overset{L}{\int }}\left(\sqrt{L^2+y^2}-y\right)\end{array}$

(substituting the given values )

$$\begin{gathered} =\left(9.0\times {10}^{9}N{m}^2/C^2\right)\left(49.9\times {10}^{-12}C/m^2\right)\left(\sqrt{{\left(0.10m\right)}^2+{\left(0.0356m\right)}^2}-0.0356m\right) \\ =31.68\times {10}^{-3}V \\ =31.68mV \end{gathered}$$

Hence, the value of the electric potential is 31.68mV.

The electric field at the point can be given using equation (iii) and equation (a) as follows:

$\begin{array}{rcl}{E}_y& =& -\frac{c}{4\pi {\epsilon }_0}\frac{\partial }{\partial y}\left(\sqrt{L^2+y^2}-y\right)\\ & =& \frac{c}{4\pi {\epsilon }_0}\left(1-\frac{y}{\sqrt{L^2+y^2}}\right)\\ & =& \left(9.0\times {10}^{9}N{m}^2/C^2\right)\left(49.9\times {10}^{-12}C/m^2\right)\left(-1\frac{0.0356m}{\sqrt{{\left(0.1m\right)}^2+{\left(0.0356m\right)}^2}}\right)\\ & =& 0.298N/C\end{array}$

Hence, the value of the field is 0.298N/C.

The electric field component E_x at P₂ cannot be found by using the result of part (a). This is because $V=\frac{c}{4\pi {\epsilon }_0}\left(\sqrt{L^2+y^2}-y\right)$ is a function of y, not a function of x.