Question

A thin rod lies on the x axis with one end atand the other end at$\mathbf{-}\mathbf{A}$, as shown in Figure 15.51. A charge of$\mathbf{-}\mathbf{Q}$
is spread uniformly over the surface of the rod. We want to set up an integral to find the electric field at location $<0,Y,0>$due to the rod. Following the procedure discussed in this chapter, we have cut up the rod into small segments, each of which can be considered as a point charge. We have selected a typical piece, shown in red on the diagram

Answer using the variables $\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{dx}\mathbf{,}\mathbf{A}\mathbf{,}\mathbf{Q}$as appropriate. Remember that the rod has charge$\mathbf{-}\mathbf{Q}$. (a) In terms of the symbolic quantities given above and on the diagram, what is the charge per unit length of the rod? (b) What is the amount of charge$\mathbf{dQ}$on the small piece of length$\mathbf{dx}$? (c) What is the vector from this source to the observation location? (d) What is the distance from this source to the observation location? (e) When we set up an integral to find the electric field at the observation location due to the entire rod, what will be the integration variable?

Short answer

a)$\frac{-Q}{2A}$

b)role="math" localid="1655981162881" $\frac{-Qdx}{2A}$

c)

d)$\sqrt{x^2+y^2}$

e) x

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The given starting position of the rod is,$I_1=A$
• The given end position of the rod is,$I_2=-A$
• The location of the electric field is,
• The charge of the rod is,$E=-Q$

Significance of the electric field and equation of charge per unit length

The electric field is a region which helps a charged particle to induce force on another charged particle.

The equation of the charge per unit length is expressed as,

$q=\frac{E}{a}$…(1)

Here,$E$ is the charge of the rod and is the total area of the rod.

Determination of the charge per unit length of the rod

(a)

Rewrite equation (1) as follows,

$q=\frac{E}{I_1+I_2}$

Here,$I_1$ and$I_2$ are the area at the starting point and end point respectively

For$E=-Q,I_1=A$,and$I_2=-A$.

$$\begin{gathered} q=\frac{-Q}{A-\left(-A\right)} \\ =\frac{-Q}{2A} \end{gathered}$$…(2)

Thus, the change per unit area of the rod is$\frac{-Q}{2A}$.

Determination of the amount of charge on the small piece of length

(b)

The small amount of the charge$dQ$on small piece of length$dx$, is expressed by using equation (2) as follows,

$$\begin{gathered} dq=\frac{-Q}{2A}dx \\ =-\frac{Qdx}{2A} \end{gathered}$$…(3)

Thus, the amount of the charge $dQ$on the small piece of length $dx$is$-\frac{Qdx}{2A}$$-\frac{Qdx}{2A}$.

Determination of the vector from the source to the observation location

(c)

As the electric field is at the locationand as it is in the$xy$ plane, so there will not be any$z$ coordinate. However, as the length of the$-x$ rod starts from the direction, hence, the vector fromthe source to the observation locationis .

Thus, the vector from this source to the observation location is.

Determination of the distance from the source to the observation location

(d)

The equation of the distance of the source to the observation location is expressed as,

$d=\sqrt{a^2+b^2+c^2}$

Here,$a$ , $b$and $c$are the values of the$x$ ,$y$ , and$z$ coordinate respectively.

For $a=-x$and$b=y$,

$$\begin{gathered} d=\sqrt{{\left(-x\right)}^2+y^2} \\ =\sqrt{x^2+y^2} \end{gathered}$$

Thus, the distance from the source to the observation location is$\sqrt{x^2+y^2}$.

Determination of the integration variable

(e)

According to the equation (3), as $dx$is the change of the length, then if an integral is set up to find the electric field at the observation location due to the entire rod, then the integration variable will be$x$ .

Thus, the integration variable will be $x$.