Question

A hollow glass tube A thin-walled hollow circular glass tube, open at both ends, has a radius radius R and length L (see the figure). The axis of the tube lies along the x axis, with the left end origin. The outer sides are rubbed with silk and acquire a net positive change distributed uniformly.

Answer 1

surface charge density of glass tube is,

$\dpi{80} \fn_jvn \sigma = \frac{Q}{2 \pi RL}$

now electric field at point P is given by,

The component dEx of that electric field along the direction of the axis perpendicular to the plane of the ring is,

dE_x = dE cos

$\dpi{80} \fn_jvn dE_{x} = dE \left ( \frac{b}{r} \right)$

$\dpi{80} \fn_jvn dE_{x} = k\frac{Q}{r^2} \left ( \frac{b}{r} \right)$

whrer r is ,

$\dpi{80} \fn_jvn r = \left [ \left ( \frac{R}{2} \right )^2 + (\omega -x)^2 \right ]^{\frac{1}{2}}$

and $\dpi{80} \fn_jvn b= (\omega -x)$

hence,

$\dpi{80} \fn_jvn E_{x} = k \int_{0}^{L} \frac{ \omega -x}{[(\frac{R}{2})^2+ (\omega -x)^2]^{\frac{3}{2}}} \left ( \frac{Q}{2 \pi RL} \right ) 2 \pi R dx$

$\dpi{80} \fn_jvn E_{x} = \frac{kQ}{L} \int_{0}^{L} \frac{ \omega -x}{[(\frac{R}{2})^2+ (\omega -x)^2]^{\frac{3}{2}}} dx$

by integrating,

$\dpi{80} \fn_jvn E_{x} = \frac{kQ}{L} \left [ \frac{2}{ \sqrt{[ 4(x- \omega )^2 + R^2 ]}} \right ] _{0}^{L}$

$\dpi{80} \fn_jvn E_{x} = \frac{kQ}{L} \left [ \frac{2}{ \sqrt{[ 4(L- \omega )^2 + R^2 ]}} - \frac{2}{ \sqrt{[ 4(- \omega )^2 + R^2]}} \right ]$

by pluging the given values, of L , R and $\dpi{80} \fn_jvn \omega$

$\dpi{80} \fn_jvn E_{x} = \frac{(8.99 \times 10^{9} Nm^2/C^2)(9 \times 10^{-9}C)}{0.05m} \left [ \frac{2}{ \sqrt{[ 4(0.05m- 0.1m )^2 + (0.03m)^2 ]}} - \frac{2}{ \sqrt{[ 4(- 0.1m )^2 + (0.03m)^2]}} \right ]$

$\dpi{80} \fn_jvn E_{x} = 8990Nm/C \left [ \frac{2}{ \sqrt{ 0.0109}} - \frac{2}{ \sqrt{0.0409}} \right ]\frac{1}{m}$

$\dpi{80} \fn_jvn E_{x} = 8990 \left [ 19.15- 9.88 \right ] \frac{N}{C}$

$\dpi{80} \fn_jvn E_{x} = 83313.02 \, \frac{N}{C}$