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.

b) In the figure there are three infinite long thin sheets having surface charge density +2σ, -2σ and +σ respectively. Give the magnitude and direction of electric field at a point to the left of sheet of charge density +2σ and to the right of sheet of charge density +σ.

Answer 1

(a) Gauss's law in electrostatics: 

The surface integral of electrostatic field \(\vec{E}\) produced by any sources over any closed surface s enclosing a volume V in vacuum i.e., total electric flux (ϕ_E) over the closed surface s in vacuum is \(\frac{1}{\epsilon_o}\)times the total charge (Q) contained inside s, i.e.,

ϕE = \(\int_s\vec{E}.\vec{ds}=\frac{Q}{\epsilon_o}\)

dϕ = \(\vec{E}.\vec{ds}\)

= \(\frac{Kq}{r^2}\hat{r}.\vec{ds}\)

= \(\frac{Kq}{r^2}\hat{r}.{ds}.d\hat{s}\)

∴ \(\int_s\) dϕ = \(\int_s\)Kq \(\big(\frac{ds}{r^2}\big)\hat{r}.d\hat{s}\)

ϕ = \(\int_s\) Kq\(\big(\frac{ds}{r^2}\big).1\)  as \(\big(\hat{r}.d\hat{s}=|\hat{r}||d\hat{s}|\cos0^o=1\big)\)

= Kq\(\int_s\)\(\frac{ds}{r^2}\) = Kq.\(\frac{4\pi r^2}{r^2}\) 

= Kq.4π

ϕ = \(\frac{1}{4\pi \epsilon_o}\).Kq.4π = \(\frac{q}{\epsilon_o}\)

 ϕ = \(\frac{Q}{\epsilon_o}\) (i.e., Independent of shape and size)

(b) 

\(\vec{E_A}=\frac{2\sigma}{2\epsilon_o}(-\hat{i})+\frac{2\sigma}{2\epsilon_o}(\hat{i})+\frac{\sigma}{2\epsilon_o}(-\hat{i})\)

= \(\frac{\sigma}{2\epsilon_o}(-\hat{i})\)

\(\vec{E_D}\) = \(\frac{\sigma}{2\epsilon_o}(\hat{i})-\frac{2\sigma}{2\epsilon_o}(\hat{i})+\frac{2\sigma}{2\epsilon_o}(\hat{i})\)

= \(\frac{\sigma}{2\epsilon_o}(\hat{i})\)

Answer 2

(a) Statement of Gauss law 

Proof of outward flux due to a point charge Q ,in vacuum within gaussian surface, is independent of its size and shape 

(b) Net electric field towards left=\(\frac{\sigma}{\epsilon}\) left 

Net electric field towards right=\(\frac{\sigma}{\epsilon}\) right