Question

a) State Gauss’s law in electrostatics. Show that with help of suitable figure that outward flux due to a point charge Q, in vacuum within gaussian surface, is independent of its size and shape.

Answer 1

Answer:

The surface integral of electrostatic field E⃗ E→ produced by any sources over any closed surface s enclosing a volume V in vacuum i.e., total electric flux (ϕE) over the closed surface s in vacuum is 1ϵo1ϵotimes the total charge (Q) contained inside s, i.e., ϕE = ∫sE⃗ .ds→=Qϵo∫sE→.ds→=Qϵo dϕ = E⃗ .ds→E→.ds→ = Kqr2r^.ds→Kqr2r^.ds→ = Kqr2r^.ds.ds^Kqr2r^.ds.ds^ ∴ ∫s∫s dϕ = ∫s∫sKq (dsr2)r^.ds^(dsr2)r^.ds^ ϕ = ∫s∫s Kq(dsr2).1(dsr2).1 as (r^.ds^=|r^||ds^|cos0o=1)(r^.ds^=|r^||ds^|cos⁡0o=1) = Kq∫s∫sdsr2dsr2 = Kq.4πr2r24πr2r2 = Kq.4π ϕ = 14πϵo14πϵo.Kq.4π = qϵoqϵo ϕ = QϵoQϵo (i.e., Independent of shape and size)Read more on Sarthaks.com - https://www.sarthaks.com/969050/state-gausss-electrostatics-show-that-with-suitable-figure-that-outward-flux-point-charge