Electric field due to an Infinite Wire

I am studying electrostatics from Griffiths book, and was seeing the calculation of electric field due to an infinite wire by Gauss's Law, by taking a Gaussian cylinder.

1. My doubt is, why we are finding electric field at a distance z from the wire and not along the axis of a the wire.

Question might not make sense to many readers, but I am facing a hard time understanding the geometry, conceptual problem and visualizing where and where not it doesn't makes sense to find electric field.

1. And my second doubt is, I do not understand or unable to imagine, how symmetry cancels the contribution of end caps of the Gaussian cylinder and electric field is only due to the curved part of the cylinder. If someone can help me show diagrammatically would be easy to understand it.

Note- For the second part of symmetry where two caps of cylinder makes no contribution in electric field, is it because that two ends are so far away that at a distance z they makes no sense as their contribution would be negligible,

                                  or


The two ends have area vector in opposite direction producing equal amount of electric field and thus cancelling each other fields.

• Related : Why don't stationary charge feel force from a current carrying wire?. Equations (A-01), (A-02),(A-03) and Figure-03. Commented Nov 5, 2022 at 9:25
• Have you tried using the concept of electrostatics to show there isn't any field directed through the caps in the first place? Commented Nov 5, 2022 at 14:36

First, let me try to answer your second question.

And my second doubt is, I do not understand or unable to imagine, how symmetry cancels the contribution of end caps of the Gaussian cylinder and electric field is only due to the curved part of the cylinder. If someone can help me show diagrammatically would be easy to understand it.

By cylindrical symmetry, we have: $$\mathbf{E}(\mathbf{r}) = E(\rho) \, \hat{\mathbf{\rho}}$$ Now, the Gauss law states that $$\oint_S \mathbf{E} \cdot d\mathbf{a} = \frac{Q_{enc}}{\varepsilon_0}$$ where $$S$$ is the gaussian surface of our choice. If we choose a cylinder of radius $$\rho$$ and length $$L$$, then we have (writing $$d\mathbf{a}$$ in cylindrical coordinates) $$\oint_S \mathbf{E} \cdot d\mathbf{a} = \int_0^{2\pi} \int_0^L (E(\rho) \, \hat{\mathbf{\rho}}) \cdot (\rho \, d\phi \, dz \, \hat{\mathbf{\rho}}) \; + \; 2 \int_0^r \int_0^{2\pi} (E(\rho) \, \hat{\mathbf{\rho}}) \cdot (\rho \, d\rho \, d\phi \, \hat{\mathbf{z}})$$ $$\hat{\mathbf{\rho}}$$ and $$\hat{\mathbf{z}}$$ make up an orthonormal basis, thus $$\hat{\mathbf{\rho}} \cdot \hat{\mathbf{z}} = 0$$, and we have $$\int_0^{2\pi} \int_0^L (E(\rho) \, \hat{\mathbf{\rho}}) \cdot (\rho \, d\phi \, dz \, \hat{\mathbf{\rho}}) \; + \; 2 \int_0^r \int_0^{2\pi} (E(\rho) \, \hat{\mathbf{\rho}}) \cdot (\rho \, d\rho \, d\phi \, \hat{\mathbf{z}}) = \int_0^{2\pi} \int_0^L (E(\rho) \, \hat{\mathbf{\rho}}) \cdot (\rho \, d\phi \, dz \, \hat{\mathbf{\rho}}) = E(\rho) \rho \int_0^{2\pi} \int_0^L d\phi \, dz = 2 \pi \rho L \, E(\rho)$$ The last equality holds because we do not integrate over $$\rho$$ and $$E$$ only depends on $$\rho$$.

In terms of visual intuition, you can think of it this way: the infinite cylinder is a "sum" of infinite number of point charges positioned in a line. The line is infinitely long, thus each field component in $$z$$ direction is cancelled by "neighbouring" point charges. As a result, the field depends only on the distance from the axis, which is just another way of saying that the field is cylindrically symmetric.

My doubt is, why we are finding electric field at a distance z from the wire and not along the axis of a the wire.

As you finish the calculation, you'll notice that the field is proportional to $$\frac{1}{\rho}$$. Thus, it is not sensible to ask about the field on the axis, because as $$\rho \rightarrow 0$$, $$E \rightarrow \infty$$.

• Hey @sbaginski, can you a little bit explain me why because of cylindrical symmetry there is no electric field in phi and z direction. Commented Nov 6, 2022 at 13:04
• Also if we skip and don't talk about field in radial direction, what about field in z direction, if one wants to compute there. Commented Nov 6, 2022 at 13:06
• At every point along the wire, there is equal amount of charge to positive z and negative z direction this means that the component in the z direction cancels out. Similarly there is no charge that can even cause any component in the phi direction, with our chosen coordinate system, you could define a new coordinates that have a phi direction but it would.just be the same field in a different position. Commented Nov 6, 2022 at 13:09
• Also the last comment, do you mean finding the E field on the axis of the wire? If so it is undefined Commented Nov 6, 2022 at 13:10
• The absence of z component is the result of the components cancelling out, due to the wire consisting of infinitely many point charges. As for the phi component, recall that the field of a point charge only points away from it (or towards it if the charge is negative). Thus there is no phi component, which remains true after we sum up all the charges. Moreover, in electrostatics the curl of electric field is zero as can be seen from $\nabla \times E = - \frac{\partial B}{\partial t}$ (the magnetic field is constantly zero). You can think of it as electric field not circulating anywhere. Commented Nov 6, 2022 at 13:13