### Tensile force of a linear charge

 Academic exercises often assume the existence of an external electric field caused by a straight line of constant linear charge density.This analysis derives the axial tension which would be exerted on a finite straight charged wire by this hypothetical charge distribution.  Since all charges push out against all other charges in both directions, the charges at each end of the conductor experience an outward force equal to the sum of the forces from all the other charges.  If the field intensity at the end is less than the corona potential, the conductor retains the charge with tension.If lambda designates the linear charge density which is assumed to be constant, the charge contained in a small arbitrary length is.$\Delta q=\lambda \Delta x$The end charge will be designated with the subscript zero and each adjacent charge will be designated by a subscript incremented by one.All delta-xs being equal, all delta-qs are equal and the distance from charge zero to charge n is:$x_{n}=n\Delta x$The force contribution on charge zero from charge n is:$\Delta F_{n}=\frac{1}{4\pi \epsilon _{0}}\frac{\Delta q^{2}}{x_{n}^{2}}=\frac{1}{4\pi \epsilon _{0}}\frac{\lambda ^{2}}{n^{2}}$As delta-x is made arbitrarily small, maximum n approaches infinity and the total force on q0 becomes:$F=\frac{\lambda ^{2}}{4\pi \epsilon _{0}}\sum_{n=1}^{\infty }\frac{1}{n^{2}}$Since$\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6}$The tension produced by charge density lambda is: $F=\frac{\pi \lambda ^{2}}{24\epsilon _{0}}$With lambda in coulombs per meter, the tensile force in newtons is:$F=(14.8\times 10^{9})\lambda ^{2}$An argument can be made, however, that a constant charge density is not possible in a straight line.  While the total force is bounded, the total energy is not.  The energy between two delta-qs is proportional to 1/x, making the total energy proportional to the divergent sum:$\sum_{n=1}^{\infty }\frac{1}{n}=\infty$A more intuitive argument against the existence of a constant linear charge density in a finite straight conductor is that each differential charge must experience an equal force from each direction.  (If there were a force imbalance, the charge would move.)  Charges close to one end of the conductor must have a higher linear density in order to match the force from the charges toward the far end.  e.g. consider a charge located 3/4 of the length from the left end, and 1/4 length from the right end of a straight conductor.  If lambda were constant, the force from the 1/4 length to the right would be matched by the first 1/4 length of charges to the left which only gets us to the center of the conductor.  The 1/2 length of charges from the center to the left end would cause a force imbalance, moving the charge to the right.