Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

#School

(a) Using Gauss law, derive expression for electric field due to a spherical shell of uniform charge distribution $\sigma$ and radius $R$ at a
point lying at a distance x from the centre of shell, such that

(i) $0< x< R,$ and

(ii) $X> R$

(b) An electric field is uniform and acts along $+x$ direction in the region of positive $x$ . It is also uniform with the same magnitude
but acts in $-x$ direction in the region of negative $x$ . The value of the field is $E=200\; N/C$ for $x> 0$ and $E=-200\; N/C$ for $x< 0$ . A right circular cylinder of length $20\; cm$ and radius $5\; cm$ has its centre at the origin and its axis along the x-axis so that one flat face is at $x=+10\; cm$ and the other is at $x=-10\; cm$ . Find :

(i) The net outward flux through the cylinder.

(ii) The net charge present inside the cylinder.

Answers (1)

i) $0<x<R$

Inside a spherical shell, the charge =0, therefore the electric field $E= 0$

(ii) $x>R$

Consider a point P outside the shell at a distance r. Consider a spherical Gaussian surface of radius r.

Then by Gauss law :

Flux enclosed by the surface

$\phi =\oint E.ds=\frac{q}{\varepsilon _{o}}$

$E\times 4\pi r^{2}=\frac{q}{\varepsilon _{o}}$

$E = \frac{q}{4\pi \varepsilon _{o}r^{2}}$

Here

$r=x$

and

$q=\sigma\times4\pi R^2$

therefore

$E=\frac{\sigma R^2}{\epsilon_0 x^2}$

b)

i) The net outward flux=2EA

$\\\phi=2EA=2\times200\times3.14\times(0.05)^2\\=3.14\frac{Nm^2}{C}$

ii) The net charge present inside the cylinder

$=\epsilon_0\times\phi=8.854\times10^{-12}\times3.14=2.78\times10^{-11}C$

Posted by

Safeer PP

View full answer

Similar Questions

Latest Question