# How to find direction of Electric field lines due to infinite charge distribution?

I saw in my textbook while calculating Electric field due to a uniformly charged conducting solid sphere the direction of electric fields were taken radially outward. How can you justify this? There are many charges at surface (bcoz conductor) how can you just say that the direction is radially outward Not only in this but in many situations like Electric field due to charged conducting cylinder, the field lines were taken perpendicular to the curved surface and parallel to the plane surfaces and also while calculating Electric field due to a solid non conducting cylinder (inside the cylinder), the direction is taken to be normal to the imagined gaussian cylinder inside the real inside.How can just ignore the charges outside the gaussian cylinder, they are uniformly distributed in the volume, how can you except they produce field lines normal to the surface. Any help would be appreciated. I mean how do we find the direction of field lines in such infinite charge distribution

Of course the correct answer would be to solve Maxwells equations using the divergence theorem. But I think the most intuitive answer that one can give is that a sphere is completely rotationally symmetric. No matter how you turn a sphere around its origin, the charge distribution looks the same. Thus of course the field lines have to look the same, too.

The only way to have uniform field lines around a spherically symmetric object is if you have field lines that point out radially. I hope that this answers the question you have sufficiently, because you seem to be looking for a intuitive understanding of the reason.

• Thank you very much for answering . Yes I was looking for intutive understanding of the reason . Where I can get the information about how to derive the direction of electric fields with the help of divergence theorem – user208763 May 6 '19 at 13:39
• See for example this video for an example of using the divergence theorem for just that purpose. – Iconstrife May 7 '19 at 12:58
• Thank you very much – user208763 May 9 '19 at 15:53

The shape of electric field lines comes from the expression of electric field. For a sphere, electric field vector is given as $$\vec E=\frac {kq} {r^3} \vec r$$ (the expression can be derived from E=-dV/dr). So the field lines are radial. For a cylinder, the electric field vector comes out to be along the radius vector of cylinder, perpendicular to its axis. For sphere, V=kq/r. And, $$\vec E=-\frac {\partial V}{\partial x} \vec i-\frac {\partial V}{\partial y} \vec j-\frac {\partial V}{\partial z} \vec k$$ Now ,$$\frac {\partial V}{\partial x}=\frac {\partial }{\partial x}({\frac {kq } {\sqrt{x^2+y^22+z^2}}}) = -\frac {kqx} {({x^2+y^2+z^2})^{3/2}}$$. So, plugging in the values, you get, $$\vec E=\frac {kq} {({x^2+y^2+z^2})^{3/2}}(x i+yj +zk)$$

• Thank you very much. can you give me a link for the PDF or any other source to know how to derive the shape of electric field lines from E=-dv/dr – user208763 May 6 '19 at 13:36
• You can see my edited answer for that – Tojrah May 6 '19 at 15:20
• Thank you once again – user208763 May 7 '19 at 3:00