Suppose we have two very long parallel wires of radius $a$ and distance between their axes $d$ ($d>>a$). The first wire if is carrying charge of linear charge density $\alpha$ and is insulated by a layer of linear dielectric, whose width is $a$ and relative permittivity $\epsilon_r$. Second wire is carrying charge of linear charge density $-\alpha $. Space between the wires is vacuum. Calculate force per unit length of the first wire on the second one.

First of all, the solution provided in the textbook doesn't include $\epsilon_r$ at all, which was weird but i decided to try and find my own solution. I tried using the method of virtual works, as follows:

Let $F'$ be the force per unit length we are looking for. Suppose we are keeping $|\alpha| $ constant for both wires and due to electric force, one wire will move by $dx$. $$F' = \frac{dW_e}{dx}$$ $$We=\frac{C'U^2}{2}=\frac{\alpha^2}{2C'}$$ where $C'$ is capacitance per unit length. The problem here is calculating capacitance per unit length ( i.e. calculating voltage between wires, since $\alpha = C'U$).In order to calculate the voltage, we are required to calculate electric field at an arbitrary point between the wires. Suppose this point is at distance $r$ from the axis of the first wire. By using generalizes Gauss' law, we find the electric field of the first wire at that point i.e. $E=\frac{\alpha}{4\pi\epsilon_0\epsilon_ra}$. Then we calculate field from the second wire i.e. $\frac{\alpha}{2\epsilon_0r\pi}$. Since there fields are in the same direction we add their magnitudes.Then the voltage is given by: $$E=\frac{\alpha}{2\epsilon_0\pi}\int_a^{d-a}(\frac{1}{2\epsilon_rr}+\frac{1}{r})dr$$. Solving this doesn't get me even close to the solution which appears to be $$F'=\frac{\alpha^2}{2\pi\epsilon_0d}$$


1 Answer 1


Force per unit length would be given by $$F=\alpha E.$$ For an infinite line charge the electric field at a distance $d$ is, by Gauss' Law, $$E=\frac{\alpha}{2\pi \epsilon_{0} d}.$$ The dielectric is made of dipoles, so you should be able to figure out why it makes no difference to the field outside the wire. And the two wires are far enough apart that we can ignore polarization created by the other wire's field.

  • $\begingroup$ I think i understand. Since dielectric is outside of the sphere and does not affect field source ( like the case with spherical capacitor for example ) but won't dielectric molecules still be polarized? It appears dielectric presence doesn't alter propagation of the field. But what happens at an arbitrary point at distance $r (a<r<2a)$ from the first wire. $\endgroup$
    – user16688
    Commented Nov 29, 2014 at 7:51
  • 1
    $\begingroup$ I think you mean cylinder, not sphere. Dielectric molecules contain equal amounts of positive and negative charge, so the total charge enclosed will not be affected if our Gaussian surface is outside the dielectric material. If the two wires are close ($d$ is not much larger than $a$, then the second wire's electric field would spoil the cylindrical symmetry of the first, by affecting the polarization of the dielectric, and it would not be easy to apply Gauss' Law. That's why the question stated $d>>a$. $\endgroup$
    – G. Paily
    Commented Nov 29, 2014 at 23:56

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