(a) Figure 24-42ashows a non-conducting rod of length L = 6.00cmand uniform linear charge density $\mathit{\lambda }\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{68}\mathbf{p}\mathbf{C}\mathbf{/}\mathbf{m}\mathbf{)}$. Assume that the electric potential is defined to be V = 0at infinity. What is Vat point Pat distance d = 8.00cmalong the rod’s perpendicular bisector? (b) Figure 24-42bshows an identical rod except that one half is now negatively charged. Both halves have a linear charge density of magnitude $\mathbf{3}\mathbf{.}\mathbf{68}\mathit{p}\mathit{C}\mathbf{/}\mathit{m}$. With V = 0at infinity, what is the net electric potential at the

VatP?

1. Distance from point P to the rod is d = 0.08m
2. Length of rod, L = 0.06m
3. Linear charge density, $\lambda =3.68\times {10}^{-12}pC/m$
4. The electric potential is V = 0 at infinity.

Using the concept of linear charge density, we get the value of the electric potential of the line charges on the thin conducting rod by substituting the given values in the expression of the potential of the system of the charges.

Formulae:

Electric potential due to point charge at point P,$dV=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{dq}{r}$ (i)

The linear density of a body,$\lambda =\frac{dq}{dx}$ (ii)

Let small element is at a distance x from the origin and the distance between point P and small element is given by, $r=\sqrt{x^2+d^2}$ (iii)

Considering a small element along the axis of the thin rod, then the charge on this small element is given using equation (ii) as:$dq=\lambda dx$

Now, substituting this value and equation (iii) in equation (i), we can get the value of the electric potential as follows:

role="math" localid="1662699871539" width="155" height="48">dV=14πε0λdxx2+d2

Now, this above potential expression is integrated to get the equation of the electric potential of a thin rod as follows:

role="math" localid="1662700089999" $\begin{array}{rcl}V& =& \underset{0}{\overset{L}{\int }}\frac{1}{4\pi {\epsilon }_0}\frac{\lambda dx}{\sqrt{x^2+d^2}}\\ & =& \frac{1}{4\pi {\epsilon }_0}\underset{0}{\overset{L}{\int }}\frac{\lambda dx}{\sqrt{x^2+d^2}}\\ & =& \frac{\lambda }{4\pi {\epsilon }_0}{\left[In\left(x+{\left(x^2+d^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\right]}_0^L\\ & =& \frac{\lambda }{4\pi {\epsilon }_0}In\left[\frac{L+{\left(L^2+d^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{d}\right].................\left(a\right)\end{array}$

Here L is the length of the rod and d is distance to the given point P from the rod.

Electric potential due to each half at the given point P is given by replacing L with L/2 in equation (a) as follows:

role="math" localid="1662700228718" $V=\frac{\lambda }{4\pi {\epsilon }_0}In\left[\frac{{\displaystyle \frac{L}{2}}+{\left({\displaystyle \frac{L^2}{2^2}}+d^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{d}\right]$

Electric potential due to both halves of the rod at point P is given using the above equation as follows:

$\begin{array}{rcl}V& =& \frac{\lambda }{4\pi {\epsilon }_0}In\left[\frac{{\displaystyle \frac{L}{2}+{\left(\frac{L^2}{2^2}+d^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{d}\right]\\ & =& \frac{2\left(3.68\times {10}^{-12}C/m\right)}{4\mathrm{\pi }\left(8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right)}In\left[\frac{{\displaystyle \frac{0.06m}{2}}+\sqrt{{\displaystyle \frac{{\left(0.06m\right)}^2}{4}}+{\left(0.08m\right)}^2}}{0.08m}\right]\\ & =& 2.42\times {10}^{-2}V\\ & =& 24.2mV\end{array}$

Hence, the value of the electric potential is 24.2mV.

The potential of the first half of the rod is the same as the potential of the second half of the rod, but the charges are opposite of each other. So, the potentials of both halves are equal and opposite.

Hence, the net electric potential at P, is 0mV