91影视

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

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

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

$$\begin{gathered} dE=\frac{1}{4蟺{蔚}_0}\frac{dQ}{r^2} \\ =\frac{1}{4蟺{蔚}_0}\frac{\frac{Q}{L}d{x}^{\mathrm{鈥�}}}{r^2} \\ =\frac{Q}{4蟺{蔚}_{0}L}\frac{d{x}^{\mathrm{鈥�}}}{\left(x+\frac{a}{2}+L鈭�x^{\mathrm{鈥�}}\right)} \end{gathered}$$

Now integrating the above equation, we get:

$$\begin{gathered} {鈭�}_0^l鈥�dE={鈭�}_0^1鈥�\frac{Q}{4蟺{蔚}_{0}L}\frac{d{x}^{\mathrm{鈥�}}}{\left(x+\frac{a}{2}+L鈭�x^{\mathrm{鈥�}}\right)} \\ =\frac{Q}{4蟺{蔚}_{0}L}{鈭�}_0^1鈥�\frac{d{x}^{\mathrm{鈥�}}}{\left(x+\frac{a}{2}+L鈭�x^{\mathrm{鈥�}}\right)} \\ =\frac{Q}{4蟺{蔚}_{0}L}\left(\frac{1}{\left(x+\frac{a}{2}+L鈭�L\right)}鈭�\frac{1}{\left(x+\frac{a}{2}+L鈭�0\right)}\right) \\ =\frac{Q}{4蟺{蔚}_{0}L}\left(\frac{1}{\left(x+\frac{a}{2}\right)}-\frac{1}{\left(x+\frac{a}{2}+L\right)}\right) \end{gathered}$$

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:

$$\begin{gathered} dF=EdQ \\ =\frac{Q}{4蟺{蔚}_{0}L}\left(\frac{1}{\left(x+\frac{a}{2}\right)}鈭�\frac{1}{\left(x+\frac{a}{2}+L\right)}\right)dQ \end{gathered}$$

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

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

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

$$\begin{gathered} 鈭�dF=F \\ =\frac{Q}{4蟺{蔚}_{0}L}{鈭�}_{\frac{a}{L}}^{\frac{a}{2}+L}鈥�\left(\frac{1}{{\left(x+\frac{a}{2}\right)}^{+}}+\frac{1}{\left(x+\frac{a}{2}+L\right)}\right)dQ \\ =\frac{Q^2}{4蟺{蔚}_0{L}^2}{\left(\ln \left(\frac{x+\frac{a}{2}}{x+\frac{a}{2}+L}\right)\right)}_{\frac{a}{2}}^{\frac{a}{2}+L} \\ =\frac{Q^2}{4蟺{蔚}_0{L}^2}\ln \left(\frac{(a+L{)}^2}{a(a+2L)}\right) \end{gathered}$$

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

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

Now taking a as a common factor we get:

$$\begin{gathered} F=\frac{Q^2}{4蟺{蔚}_0{L}^2}\ln \left(\frac{a^2(1+L{)}^2}{a^2\left(1+2\frac{L}{a}\right)}\right) \\ =\frac{Q^2}{4蟺{蔚}_0{L}^2}\left(2\ln \left(1+\frac{L}{a}\right)鈭�\ln \left(1+2\frac{L}{a}\right)\right) \end{gathered}$$

Now using the given expansion, we solve:

$$\begin{gathered} F=\frac{Q^2}{4蟺{蔚}_0{L}^2}\left(2\ln \left(1+\frac{L}{a}\right)鈭�\ln \left(1+2\frac{L}{a}\right)\right) \\ =\frac{Q^2}{4蟺{蔚}_0{L}^2}\ln \left(\frac{a^2(1+L{)}^2}{a^2\left(1+2\frac{L}{a}\right)}\right) \\ =\frac{Q^2}{4蟺{蔚}_0{L}^2}\left(2\left(\frac{L}{a}鈭�\frac{L^2}{2a^2}+鈥�\right)鈭�\left(\frac{2L}{a}鈭�\frac{4L}{2a^2}+鈥�\right)\right) \\ =\frac{Q^2}{4蟺{蔚}_0{L}^2}\ln \left(\frac{a^2(1+L{)}^2}{a^2\left(1+2\frac{L}{a}\right)}\right) \\ =\frac{Q^2}{4蟺{蔚}_0{L}^2}\frac{L^2}{a^2} \\ =\frac{Q^2}{4蟺{蔚}_0{L}^2}\ln \left(\frac{a^2(1+L{)}^2}{a^2\left(1+2\frac{L}{a}\right)}\right) \\ =\frac{Q^2}{4蟺{蔚}_0{a}^2} \end{gathered}$$

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