Question

A thin rod lies on the x-asix with one end at -A and the other end at A, as shown in the diagram. A charge of -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 the textbook, 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.

Use the following as necessary: x, y, dx, A, Q. Remember that the rod has charge -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 dQ on the small piece of length dx?

dQ= _________

(c) What is the vector from the source to observation location?

r= _________ (this answer is a vector)

(d) What is the distance from the source to the observation location?

d= ________

(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?

answer=_________

Answer 1

Concepts and reason

The concepts required to solve this problem are the charge per unit length, electric field, and linear charge density.

First calculate the charge per unit length of the rod. After that calculate the charge on small segment, then calculate the vector from the source to observation location and the distance from the source to observation location. Finally calculate the integration variable to find the electric field.

Fundamentals

The electric force per unit charge is called electric field. The direction of field is the direction of force exerted on a positive test charge. The electric field on positive charge is radially outward and on negative charge is radially inwards.

The charge per unit length is defined the ratio of total charge to the total length.

${\rm{Charge per unit length of rod}} = \frac{{{\rm{total charge}}}}{{{\rm{total length}}}}$

The expression of electric field for point charge is,

$E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}}$

Here, $Q$ is charge, $r$ is distance, and ${\varepsilon _0}$ is permittivity of free space.

(a)

The image shows a thin rod of length $2A$ with total charge $- Q$ . Consider a small length element $dx$ at the distance of $x$ from the origin. The angle of electric field from the $y$ -axis at point $P$ is $\theta$ .

Calculate the charge per unit length of the rod.

Charge per unit length of rod is calculated by using the formula,

$\lambda = \frac{{{\rm{total charge}}}}{{{\rm{total length}}}}$

Substitute $- Q$ for ${\rm{total charge}}$ and $2A$ for ${\rm{total length}}$ of the rod.

$\lambda = \frac{{ - Q}}{{2A}}$

[Part a]

Part a

(b)

Calculate the charge $dQ$ on the small piece of length $dx$ .

The charge on small segment is equal to the product of charge per unit length and length of small segment,

$dQ = \lambda dx$

Substitute $\frac{{ - Q}}{{2A}}$ for $\lambda$ .

$dQ = \frac{{ - Q}}{{2A}}dx$

(c)

Refer figure 1 and determine the vector from the source to observation location.

The location of the observer is,

$\vec O = (x,0,0)$

The location of the source is,

$\vec S = (0,y,0)$

The vector from the source to observation location is,

$\overrightarrow {OS} = \left( {{x_2} - {x_1}} \right){\rm{\hat i}} + \left( {{y_2} - {y_1}} \right){\rm{\hat j}}$

Here, ${x_1}$ and ${y_1}$ are the x and y component of the observer location and, ${x_2}$ and ${y_2}$ are the x and y components of the source location.

Substitute 0 for ${x_2}$ , x for ${x_1}$ , y for ${y_2}$ , and 0 for ${y_1}$ in the equation $\overrightarrow {OS} = \left( {{x_2} - {x_1}} \right){\rm{\hat i}} + \left( {{y_2} - {y_1}} \right){\rm{\hat j}}$ .

$\begin{array}{c}\\\overrightarrow {OS} = \left( {0 - x} \right){\rm{\hat i}} + \left( {y - 0} \right){\rm{\hat j}}\\\\ = - x{\rm{\hat i}} + y{\rm{\hat j}}\\\end{array}$

(d)

Calculate the distance from the source to observation location,

The distance from the source to observation location is calculated by using the formula,

$OS = \sqrt {{a_x}^2 + {a_y}^2}$

Here, ${a_x}$ and ${a_y}$ are the x and y components of the vector $\overrightarrow {OS}$ .

Substitute –x for ${a_x}$ and y for ${a_y}$ in equation $OS = \sqrt {{a_x}^2 + {a_y}^2}$ .

$\begin{array}{c}\\OS = \sqrt {{{\left( { - x} \right)}^2} + {{\left( y \right)}^2}} \\\\ = \sqrt {{x^2} + {y^2}} \\\end{array}$

(e)

Determine the integral to find the electric field.

The expression of electric field for point charge is,

$E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}}$

Here, $Q$ is charge, $r$ is distance, and ${\varepsilon _0}$ is permittivity of free space.

Calculate the electric field of small segment $dx$ .

Substitute $\sqrt {{x^2} + {y^2}}$ for $r$ and $dQ$ for $Q$ .

$\begin{array}{c}\\d\vec E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{dQ}}{{{{\left( {\sqrt {{x^2} + {y^2}} } \right)}^2}}}\\\\ = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{dQ}}{{\left( {{x^2} + {y^2}} \right)}}\\\end{array}$

Expression of small charge in term of linear charge density is,

$dQ = \lambda dx$

Here, $\lambda$ is linear charge density, and $dx$ is small length.

Substitute $\lambda dx$ for $dQ$ .

$d\vec E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\lambda dx}}{{\left( {{x^2} + {y^2}} \right)}}$

Linear charge density is,

$\lambda = \frac{{ - Q}}{{2A}}$ .

Here $2A$ is total length of rod and $Q$ is total charge on rod.

Substitute $\frac{{ - Q}}{{2A}}$ for $\lambda$ in equation $d\vec E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\lambda dx}}{{\left( {{x^2} + {y^2}} \right)}}$

$\begin{array}{c}\\d\vec E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( { - Q} \right)dx}}{{\left( {2A} \right)\left( {{x^2} + {y^2}} \right)}}\\\\ = \frac{1}{{8\pi {\varepsilon _0}A}}\frac{{Qdx}}{{\left( {{x^2} + {y^2}} \right)}}\\\end{array}$

The only variable in the above integral is x.

Ans: Part a

The charge per unit length of the rod $\left( \lambda \right)$ is $\frac{{ - Q}}{{2A}}$ .

Part b

The charge $dQ$ on the small piece of length $dx$ is $\frac{{ - Q}}{{2A}}dx$ .

Part c

The vector form of the source to observer location is $\left( { - x{\rm{\hat i}} + y{\rm{\hat j}}} \right)$ .

Part d

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

Part e

The integration variable is $x$ .