A thin-walled hollow circular glass tube, open at both ends, has a radius R and length L. The axis of the tube lies along the x axis, with the left end at the origin (Figure 15.58). The outer sides are rubbed with silk and acquire a net positive charge Q distributed uniformly. Determine the electric field at a location on the x axis, a distance w from the origin. Carry out all steps, including checking your result. Explain each step. (You may have to refer to a table of integrals.)

The given data can be listed below,

• The radius of the hollow circular glass tube is, R
• The length of the hollow circular glass tube is, L
• The charge on the outer side of hollow circular glass tube is, Q

A concentration of electric influence obtained from a full encirclement of charge-hosting object can be referred to as an electric field. This idea is in line with the electrical equivalence principle.

The glass tube can be considered as a cylinder with length L and radius R. the cylinder can be divided into small rings of thickness dx, charge dQ and electric field dE in the x-direction. The surface charge density of uniformly charged cylinder is given by,

$\sigma =\frac{Q}{2\pi RL}$

Here, Q is the charge on cylinder, R is the radius of cylinder and L is the length of the cylinder.

The surface of each ring is 2πRdx so the charge on the ring is given by,

$dQ=\frac{Q}{L}dx$

The electric field at point w due to the ring at position x is given by,

$\mathrm{dE}=\frac{\mathrm{K}\left(\mathrm{w}-\mathrm{x}\right)\mathrm{dQ}}{{\left[{\left({\mathrm{R}}^2+{\left(\mathrm{w}-\mathrm{x}\right)}^2\right)}^2\right]}^{3/2}}$

Here, K is the coulomb constant,$\left(w-x\right)$is the distance between two positions, R is the radius of ring.

The electric field is given by integrating both side with limits 0 to L.

$$\begin{gathered} E={\int }_0^L\frac{K\left(Q/L\right)dx\left(w-x\right)}{{\left[\left(R^2+{\left(w-x\right)}^2\right)\right]}^{3/2}} \\ =\frac{KQ}{L}{\int }_0^L\frac{\left(w-x\right)dx}{{\left[\left(R^2+{\left(w-x\right)}^2\right)\right]}^{3/2}} \end{gathered}$$

Let (w-x) = u, -dx=du.

Limit will shift to 0→w, L→ w - L.

$$\begin{gathered} E=\frac{KQ}{L}{\int }_w^{w-L}\frac{-udu}{{\left(R^2+u^2\right)}^{3/2}} \\ =-\frac{KQ}{L}{\left(-\frac{1}{\sqrt{R^2+u^2}}\right)}_w^{w-L} \\ =\frac{KQ}{L}\left[\frac{1}{\sqrt{R^2}+{\left(w-L\right)}^2}-\frac{1}{\sqrt{R^2+w^2}}\right] \end{gathered}$$

When the R approaches zero the electric field is given by,

$$\begin{gathered} E=\frac{KQ}{L}\left(\frac{1}{w-L}-\frac{1}{w}\right) \\ =\frac{KQ}{w\left(w-L\right)} \end{gathered}$$

Thus, the electric field on x-axis at w distance away from origin is .

$\frac{\mathrm{KQ}}{\mathrm{L}}\left[\frac{1}{\sqrt{{\mathrm{R}}^2+{\left(\mathrm{w}-\mathrm{L}\right)}^2}}-\frac{1}{\sqrt{{\mathrm{R}}^2+{\mathrm{w}}^2}}\right]$.