The discussion centers on the evaluation of the surface integral ##\int_v(\nabla•A)dv=\int_s(A•ds)## for large surfaces and the conditions under which terms can be neglected. It is established that the vector field ##\vec A## must approach zero at infinity faster than the surface area grows to ensure the integral evaluates to zero. The example of a point charge is used to illustrate that the field decreases as ##\frac{1}{r^2}## while the surface area grows as ##r^2sin\theta d\theta d\phi##, demonstrating that the integral is not zero if the decay rate of the field is insufficient.
PREREQUISITESPhysicists, mathematicians, and engineering students interested in advanced calculus, particularly those focusing on electromagnetism and vector field theory.
Okk got it ...thanks @OrodruinOrodruin said:The field ##\vec A## is assumed to go to zero at infinity at a rate faster than the rate at which the surface grows.
But if the rate at which the vector field ##A## goes to zero at infinity is slower than the rate at which the surface grows ,then also it would be zero ,isn't it??Orodruin said:The field ##\vec A## is assumed to go to zero at infinity at a rate faster than the rate at which the surface grows.
Ok the field decreses as ##\frac{1}{r^2}## and the spherical surface element centered at the point charge grows as ##r^2sin\theta d\theta d\phi##Orodruin said:No. You need the assumption of the field going to zero fast enough. Take the example of the field of a single point charge.
And so the integral is not zero.Apashanka said:Ok the field decreses as ##\frac{1}{r^2}## and the spherical surface element centered at the point charge grows as ##r^2sin\theta d\theta d\phi##