can gauss's law be used to calculate the electric field due to two point charges? if so, how? if not, why?

Question

i know it's a law and always true and all of that, i mean is it practical...

PEOPLE! I KNOW THE LAW! pleeeeeeeeease, just "present" the proper surface for two point charges!

Answer 1

Since Gauss's Law involves a surface integral, it usually finds its greatest use when there is enough symmetry in the problem that the integrand (normal electric field component) is constant on some surface (like a sphere for a point charge or a cylinder for a line charge). The integral, then, involves. a simple multiplication of the integrand by area. Two point charges do not have such symmetry, so Gauss's law is of little use. One must revert to Coulomb's law and apply the principle of superposition.

Answer 2

Gauss's theorem states that the surface integral of the normal component of the electric field on a closed surface is proportional to the total Electric charge inside the surface. This property, by itself, is not sufficient to calculate the electric field.

However, if there are additional properties, such as symmetry, the theorem allows us to calculate the field in a simple and ellegant way.

Examples are the field due to an infinite charge of uniform density (line, plane or volume).

Answer 3

gauss's theorm states that the total normal electric flux passing through any closed surface is 1/E0 times the net charge enclosed by the surface.
ie the charge due to even a sheet of charges can be found out.or the charge due to an infinitly long straight wire can be found
if u want more informations refer some ncert txt books

Answer 4

an hassle-free answer is to stick to Gauss's regulation. think a fastened effective fee is located interior the around cellular (choose no longer be on the middle) using Gauss regulation we are in a position to declare that this effective fee induces an equivalent volume of destructive fee on the internal floor of the around cellular and equivalent volume of effective rates on the outer floor of the around cellular. {in view that this could be a metallic floor, loose electrons are attracted by using the effective fee and are on the internal floor and hence the outer floor acquire equivalent effective fee} because of the fact the electrons pull the effective fee it is interior the around cellular in all guidelines the fee could now come to the middle of the around cellular. As for because of the fact the destructive rates that are on the internal floor are in touch, using Gauss regulation we are in a position to handle those rates as though they're centred on the middle of the sector the place there is already equivalent effective fee. hence internet fee interior is 0. In result, putting a effective fee interior a metallic hollow sphere is a similar because of the fact the expenses are on the outer floor of the hollow sphere. word that there are effective rates on the outer floor by using induction. hence the sector interior the hollow sphere is 0. that suggests the skill is a similar in the process the area interior the sector. for outdoors the sector, the sector could be calculated back by using Gauss regulation, as though the expenses are on the middle of the hollow sphere.