Question

A plastic rod $\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{m}$long is rubbed all over with wool, and acquires a charge of$\mathbf{-}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathit{C}$(Figure 15.52). We choose the center of the rod to be the origin of our coordinate system, with the x axis extending to the right, the y axis extending up, and the z axis out of the page. In order to calculate the electric field at location$\mathbf{A}\mathbf{=}\mathbf{<}\mathbf{07}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}$, we divide the rod into eight pieces, and approximate each piece as a point charge located at the center of the piece.

(a) What is the length of one of these pieces? (b) What is the location of the center of piece number 3? (c) How much charge is on piece number? (Remember that the charge is negative.) (d) Approximating piece 3as a point charge, what is the electric field at location A due only to piece 3? (e) To get the net electric field at location A, we would need to calculatedue to each of the eight pieces, and add up these contributions. If we did that, which arrow (a–h) would best represent the direction of the net electric field at location A?

Short answer

a)$0.2125\mathrm{m}$

b)$0.53125\mathrm{m}$

c)$-25\times {10}^6\mathrm{C}$

d)$7.91\times {10}^{18}\mathrm{N}/\mathrm{C}$

e) arrow j

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The length of the plastic rod is,$\mathrm{L}=11.7\mathrm{m}$
• The charge of the plastic rod is,$Q=-2\times {10}^{-8}C$
• The location A is at a distance of,

Significance of the electric field and net electric field equation

The electric field is referred to as a region that helps a charged particle to exert force on another charged particle.

The equation of the net electric field can be expressed as-

$E_1=k_e\frac{E}{{\left(d_1-d_2\right)}^2}$…(1)

Here,$k_e$ is the electric field constant, E is the charge in the piece number 3, is the location of the center of piece number 3 and$d_2$ is the distance of the location A from the origin.

Determination of the length of one piece

(a)

As the plastic rod has eight pieces, then the equation of the length of one piece is expressed as,

$I=\frac{L}{8}$

Here, $L$is the length of the rod.

For,$L=1.7\mathrm{m}$

$$\begin{gathered} I=\frac{1.7m}{8} \\ =0.2125m \end{gathered}$$

Thus, the length of one of those pieces is$0.2125\mathrm{m}$.

  Determination of the location of the center of piece number 3

(b)

As the rod’s center is located at$x=0$ , that shows that 4 pieces of the rod are on the negative side and the other 4 pieces are on the positive side of the axis. Hence, the pieces in the positive direction are piece 5, 6, 7 and 8.

The equation of the center of the piece 5 is expressed as,

$I_1=\frac{I}{2}$

For,$I=0.2125\mathrm{m}$

$$\begin{gathered} I_1=\frac{0.2125\mathrm{m}}{2} \\ =0.10625\mathrm{m} \end{gathered}$$

As the center of the piece 3 is before one piece that is piece 4 with respect to the piece 5, then the equation of the location of the center of piece number 3 is expressed as,

$I_2=I_1-2I$

Here, $I_2$is the location of the center of piece number 3

$$\begin{gathered} I_2=0.1.625\mathrm{m}+2\times 0.2125\mathrm{m} \\ =0.10625\mathrm{m}+0.425\mathrm{m} \\ =0.53125\mathrm{m} \end{gathered}$$

Thus, the location of the center of piece number 3 is $0.53125\mathrm{m}$.

Determination of the charge on piece 3

(c)

The equation of the electric field at location A is expressed as,

$E=\frac{Q}{8}$

Here, Q is the charge of the plastic rod.

For $Q=-2\times {10}^{-8}C$,

$$\begin{gathered} E=\frac{-2\times {10}^{-8}C}{8} \\ =-25\times {10}^{6}C \end{gathered}$$

Thus, the charge is on piece number 3 is$-25\times {10}^{6}C$ .

Determination of the net electric field at location A

(d)

$\mathrm{For}E=-25\times {10}^{6}C,k_e=8.99\times {10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2,{\mathrm{d}}_1=0.53125\mathrm{m}\mathrm{and}{\mathrm{d}}_2=0.7\mathrm{m}\mathrm{in}$ equation (1).

$$\begin{gathered} E_1=8.99\times {10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2\times \frac{-25\times {10}^6\mathrm{C}}{{\left(0.53125\mathrm{m}-0.7\mathrm{m}\right)}^2} \\ =8.99\times {10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2\times \frac{-25\times {10}^6\mathrm{C}}{{\left(-0.16875\mathrm{m}\right)}^2} \\ =8.99\times {10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2\times \frac{-25\times {10}^6\mathrm{C}}{0.0284{\mathrm{m}}^2} \\ =7.91\times {10}^{18}\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the electric field at location A due only to piece 3 is$7.91\times {10}^{18}\mathrm{N}/\mathrm{C}$.

Determination of the representation of the direction of the net electric field

(e)

If all the change in the energy of the pieces is being obtained, then it can be observed that all the energy will point in the middle direction as all the energy will be concentrated in a particular point and that will be the middle point and their magnitude will be zero. Hence, the arrow j best represents the direction of the net electric field at location A.

Thus, the arrow j best represents the direction of the net electric field at location A.