Question

Question: The thin plastic rod shown in Fig. 24-47 has length L = 12.0cmand a non-uniform linear charge density, $\mathit{\lambda }\mathbf{=}\mathit{c}\mathit{x}$where $\mathit{c}\mathbf{=}\mathbf{28}\mathbf{.}\mathbf{9}{\mathbf{\text{pC/m}}}^{\mathbf{\text{2}}}$.With V = 0at infinity, find the electric potential at point P₁ on the axis, at distance d = 3.00 cmfrom one end.

Short answer

Answer:

The electric potential at the point P₁ on the axis from one end is 18.6 mV.

Step-by-step solution

The given data

1. Length of the rod,L = 0.12 m
2. Non-uniform linear charge density, $\lambda =cx$where$c=28.9{\text{pC/m}}^{\text{2}}$
3. The potential is at V = 0 infinity.
4. The distance of the point on the x-axis,$d=0.03\text{m}$

Understanding the concept of the electric potential

Using the value of the charge derived from the linear charge density in the formula of the potential at the segment of the rod from one end, we can get the value of the electric potential at the given point on the x-axis.

Formulae:

The charge due to linear charge density, $dq=\lambda dx$ (i)

The electric potential due to the segment on the rod, $dV=\frac{1}{4\pi {\epsilon }_0}\left(\frac{dq}{x+d}\right)$ (ii)

Calculation of the electric potential at the point from one end  

The charge of the segment using the given data in equation (i) is given as follows:

$dq=cxdx$

The potential at point P is obtained by integrating the expression for potential due to the segment of equation (ii) by substituting the above charge value as follows:

$$\begin{gathered} V={\int }_0^L\frac{1}{4\pi {\epsilon }_0}\frac{cxdx}{\left(x+d\right)} \\ =\frac{c}{4\pi {\epsilon }_0}{\left[x-dln\left(x+d\right)\right]}_0^L \\ =\frac{c}{4\pi {\epsilon }_0}\left[L-dln\left(1+\frac{L}{d}\right)\right] \\ =\left(8.99\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(28.9\times {10}^{-12}{\text{C/m}}^{\text{2}}\right)\left[\left(0.12\text{m}\right)-\left(0.03\text{m}\right)ln\left(1+\frac{\left(0.12\text{m}\right)}{\left(0.03\text{m}\right)}\right)\right] \\ =18.6\times {10}^{-3}\text{V}\left(\frac{1\text{mV}}{{10}^{-3}\text{V}}\right) \\ =18.6\text{mV} \end{gathered}$$

Therefore, the electric potential is 18.6 mV