Consider a thin glass rod of length L lying along the x axis with one end at the origin. The rod carries a uniformly distributed positive charge Q.

At a location d > L, on the x axis to the right of the rod in Figure 15.56, what is the electric field due to the rod? Follow the standard four steps. (a) Use a diagram to explain how you will cut up the charged rod, and draw the contributed by a representative piece. (b) Express algebraically the contribution each piece makes to the electric field. Be sure to show your integration variable and its origin on your drawing. (c) Write the summation as an integral, and simplify the integral as much as possible. State explicitly the range of your integration variable. Evaluate the integral. (d) Show that your result is reasonable. Apply as many tests as you can think of

The given data can be listed below,

• The charge on the rod is,$+\mathrm{Q}$.
• The length of the rod is, L.

The electric fields aid in the visualization of charged-object interactions and the calculation of the forces that charged objects exert on one another.

To obtain the electric field, the rod is divided into small point-like pieces of length and charge. The field of each piece points towards d, which is in the positive x-direction in this case. This is shown in the figure below:

Thus, the diagram shows that the field of each piece of length $dx$and $dQ$is in positive x-direction.

The electric Filed at point due to dx directed in x-direction is given by,

$d\stackrel{\rightharpoonup }{E}=\frac{KdQ}{r^2}$ …(i)

Here, K is the coulomb’s constant whose value is $9\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2$, is the small charge at $dx$and r is the distance of piece of rod to point d.

The distance of piece of rod from point d is given by,

$$\begin{gathered} \stackrel{\rightharpoonup }{r}=\left(d,0,0\right)-(x,00) \\ =(d-x,0,0) \end{gathered}$$

The charge on the piece of rod is calculated as,

$\lambda =\frac{Q}{L}$ …(ii)

Here,$\lambda$ line charge density of rod, Q is the charge and L is the total length of the rod.

So, the charge on small piece of rod is given by,

$$\begin{gathered} dQ=\lambda dx \\ =\frac{Q}{L}dx \end{gathered}$$

Substitute the value of charge and distance in equation (i)

$$\begin{gathered} d\stackrel{\rightharpoonup }{E}=\frac{KQ}{L}\frac{dx}{{\left(d-x\right)}^2} \\ d{E}_x=\frac{KQ}{L}\frac{dx}{{\left(d-x\right)}^2} \end{gathered}$$

Thus, the electric field directed in x-direction is $d{E}_x=\frac{KQ}{L}\frac{dx}{{(d-x)}^2}$.

The total electric field of the rod is the sum of each piece is given by,

$E=\sum _l\frac{K(Q/L)d{x}_l}{{(x_l-d)}^2}$

In the infinitesimal limits the sum becomes an integral over entire rod, so the electric field is given by,

$$\begin{gathered} E={\int }_0^L\frac{K(Q/L)dx}{{(x-d)}^2} \\ =\frac{KQ}{L}{\int }_0^l\frac{dx}{{(x-d)}^2} \end{gathered}$$

Let$x-d=u$so$dx=du$

Limits will shift 0 to d and L to L-d

Substitute all these in the above expression.

$$\begin{gathered} E=\frac{KQ}{L}{\int }_d^L\frac{1}{u^2}du \\ =\frac{KQ}{L}{\overline{)\left(-\frac{1}{u}\right)}}_d^{L-d} \\ =-\frac{KQ}{L}\left(\frac{1}{L-d}+\frac{1}{d}\right) \\ =\frac{KQ}{d(d-L)} \end{gathered}$$

Thus, the electric field is positive and it’s value is $E=\frac{KQ}{d(d-L)}$.

The electric field of the rod is given by,

$E=\frac{KQ}{d(d-L)}$

When the point of observation is very far from the rod d is very greater than L so the electric field of the rod is,

$E\approx \frac{KQ}{d^2}$

This field is the electric field of the point charge that means when the rod is very far it becomes the point charge as its length is tending towards zero and its direction is positive in the positive x-direction and negative for the negative x-direction.

Thus, the results obtain are correct as the distance of the rod from point is larger or the rod is smaller leading its field to nearly equal to the point charge electric field.