Two thin rods of length L lie along the x-axis, one between and$\mathbf{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{a}\mathbf{+}\mathbf{L}$ the other between and . Each rod has positive charge Q distributed uniformly along its length. (a) Calculate the electric field produced by the second rod at points along the positive x-axis. (b) Show that the magnitude of the force that one rod exerts on the other is

(c) Show that if a>> L, the magnitude of this force reduces to. (Hint: Use the expansion $$\begin{gathered} \mathbf{dF}\mathbf{=}\mathbf{EdQ} \\ \mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{L}}\left(\frac{1}{\left(x+\frac{a}{2}\right)}‐\frac{1}{\left(x+\frac{a}{2}+L\right)}\right)\mathbf{dQ} \end{gathered}$$, valid for $\left|z\right|\mathbf{\ll }\mathbf{1}$. Carry all expansions to at least order L²/ a².) Interpret this result.

From the graph it is clear that the change in charge is given by:

$$\begin{gathered} \frac{dQ}{d{I}^{\mathrm{\prime }}}=\frac{Q}{L} \\ dQ=\frac{Q}{L}d{x}^{\mathrm{\prime }} \end{gathered}$$

To evaluate the electric field produced by the second rod we can use the following method:

$$\begin{gathered} dE=\frac{1}{4\pi {\epsilon }_0}\frac{dQ}{r^2} \\ =\frac{1}{4\pi {\epsilon }_0}\frac{\frac{Q}{L}d{x}^{\mathrm{\prime }}}{r^2} \\ =\frac{Q}{4\pi {\epsilon }_{0}L}\frac{d{x}^{\mathrm{\prime }}}{\left(x+\frac{a}{2}+L-x^{\mathrm{\prime }}\right)} \end{gathered}$$

Now integrating the above equation, we get:

$$\begin{gathered} {\int }_0^l dE={\int }_0^1 \frac{Q}{4\pi {\epsilon }_{0}L}\frac{d{x}^{\mathrm{\prime }}}{\left(x+\frac{a}{2}+L-x^{\mathrm{\prime }}\right)} \\ =\frac{Q}{4\pi {\epsilon }_{0}L}{\int }_0^1 \frac{d{x}^{\mathrm{\prime }}}{\left(x+\frac{a}{2}+L-x^{\mathrm{\prime }}\right)} \\ =\frac{Q}{4\pi {\epsilon }_{0}L}\left(\frac{1}{\left(x+\frac{a}{2}+L-L\right)}-\frac{1}{\left(x+\frac{a}{2}+L-0\right)}\right) \\ =\frac{Q}{4\pi {\epsilon }_{0}L}\left(\frac{1}{\left(x+\frac{a}{2}\right)}-\frac{1}{\left(x+\frac{a}{2}+L\right)}\right) \end{gathered}$$

To calculate the force exerts on the rod we use the following relation:

$$\begin{gathered} E=\frac{dF}{dQ} \\ dF=EdQ \end{gathered}$$

Now substituting the value of electric field we get:

$$\begin{gathered} dF=EdQ \\ =\frac{Q}{4\pi {\epsilon }_{0}L}\left(\frac{1}{\left(x+\frac{a}{2}\right)}-\frac{1}{\left(x+\frac{a}{2}+L\right)}\right)dQ \end{gathered}$$

Now from the graph we can see that the change is little which is given by:

$$\begin{gathered} \frac{dQ}{d{I}^{\mathrm{\prime }}}=\frac{Q}{L} \\ dQ=\frac{Q}{L}d{x}^{\mathrm{\prime }} \end{gathered}$$

Now, substituting from the previous calculations and integrating both sides we get:

$$\begin{gathered} \int dF=F \\ =\frac{Q}{4\pi {\epsilon }_{0}L}{\int }_{\frac{a}{L}}^{\frac{a}{2}+L} \left(\frac{1}{{\left(x+\frac{a}{2}\right)}^{+}}+\frac{1}{\left(x+\frac{a}{2}+L\right)}\right)dQ \\ =\frac{Q^2}{4\pi {\epsilon }_0{L}^2}{\left(\ln \left(\frac{x+\frac{a}{2}}{x+\frac{a}{2}+L}\right)\right)}_{\frac{a}{2}}^{\frac{a}{2}+L} \\ =\frac{Q^2}{4\pi {\epsilon }_0{L}^2}\ln \left(\frac{(a+L{)}^2}{a(a+2L)}\right) \end{gathered}$$

From the above statement we the magnitude of the electric force is give by:

$F=\frac{Q^2}{4\pi {\epsilon }_0{L}^2}\ln \left(\frac{(a+L{)}^2}{a(a+2L)}\right)$

Now taking a as a common factor we get:

$$\begin{gathered} F=\frac{Q^2}{4\pi {\epsilon }_0{L}^2}\ln \left(\frac{a^2(1+L{)}^2}{a^2\left(1+2\frac{L}{a}\right)}\right) \\ =\frac{Q^2}{4\pi {\epsilon }_0{L}^2}\left(2\ln \left(1+\frac{L}{a}\right)-\ln \left(1+2\frac{L}{a}\right)\right) \end{gathered}$$

Now using the given expansion, we solve:

$$\begin{gathered} F=\frac{Q^2}{4\pi {\epsilon }_0{L}^2}\left(2\ln \left(1+\frac{L}{a}\right)-\ln \left(1+2\frac{L}{a}\right)\right) \\ =\frac{Q^2}{4\pi {\epsilon }_0{L}^2}\ln \left(\frac{a^2(1+L{)}^2}{a^2\left(1+2\frac{L}{a}\right)}\right) \\ =\frac{Q^2}{4\pi {\epsilon }_0{L}^2}\left(2\left(\frac{L}{a}-\frac{L^2}{2a^2}+\dots \right)-\left(\frac{2L}{a}-\frac{4L}{2a^2}+\dots \right)\right) \\ =\frac{Q^2}{4\pi {\epsilon }_0{L}^2}\ln \left(\frac{a^2(1+L{)}^2}{a^2\left(1+2\frac{L}{a}\right)}\right) \\ =\frac{Q^2}{4\pi {\epsilon }_0{L}^2}\frac{L^2}{a^2} \\ =\frac{Q^2}{4\pi {\epsilon }_0{L}^2}\ln \left(\frac{a^2(1+L{)}^2}{a^2\left(1+2\frac{L}{a}\right)}\right) \\ =\frac{Q^2}{4\pi {\epsilon }_0{a}^2} \end{gathered}$$

Therefore, the magnitude of the force is:$\frac{Q^2}{4\pi {\epsilon }_0{a}^2}$