Determination of the electric field of charged object using Gauss's law

Question

when we determine the electric field of a charged rod of infinite length,we consider a circular cylinder to be the gaussian surface for convenience.In the case of a charged sheet we choose the gaussian surface to be a cylinder going through the sheet,again for convenience. yes it is easy to calculate the flux if we choose the gaussian surface to be a cylinder in the above cases?but how can we calculate the flux if we choose an arbitrary surface to be the gaussian surface?

Answer 1

You would calculate the flux $\Phi$ as \begin{equation} \Phi=\int \vec{E}\cdot d\vec{A}, \end{equation} just as you would for a sphere or a cylinder.

Gauss's law is true for any closed surface. Just take your electric field and take the dot product with the differential area $d\vec{A}$ of the surface and integrate this over the entire surface. The tricky part is that the geometry can get very ugly for arbitrary surfaces.

The reason why we usually choose familiar surfaces such as cylinders and spheres is that the differential area vectors are well-known and simple (they both point radially), but it is important to recognize that Gauss's law holds for arbitrary closed surfaces as well, even if doing the actual calculation for these surfaces is impractical.

Edit: Gauss's law is not \begin{equation} \Phi=\int \vec{E}\cdot d\vec{A}. \end{equation} This is just the definition of the electric field flux over a closed surface. Gauss's law relates the integral over a closed surface $S$ to an integral over the volume $V$ bounded by $S$. Specifically \begin{equation} \int_S \vec{E}\cdot d\vec{A}=\int_V \nabla\cdot\vec{E}\;dV. \end{equation} This result is known as the Divergence Theorem. Gauss's law makes the additional step of using Maxwell's equation for the divergence of the electric field to write \begin{equation} \int_V \nabla\cdot\vec{E}\;dV=\int_V\frac{q}{\epsilon_0}dV=\frac{Q}{\epsilon_0}, \end{equation} where $Q$ is the total charge contained in the volume $V$. Gauss's law is interesting not just because we can sometimes use it to compute the electric field of a charge configuration but because of the quantities it relates.

Gauss's law (and the Divergence Theorem) are typically the first examples that one encounters of a very general mathematical result known as the generalized Stoke's Theorem, which relates integrals over some region (which can be $n$-dimensional) to integrals over the boundary of that region (which would then be $(n-1)$-dimensional.

Answer 2

When you choose an arbitrary gaussian the flux is still given by the charge inside the surface. However it would be useless in calculating the electric field unless you are able to write $$\int\vec E\cdot d\vec A=\int EdA=E\int dA.$$ That is, Gauss law is useful when the surfaces elements are always parallel, antiparallel or perpendicular to the electric field and the magnitude of the field is constant along finite surfaces so that it can be put out of the integral.