91Ó°ÊÓ

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,$Q=-2×{10}^{-8}C$
• The location A is at a distance of,

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.

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

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

(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

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

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

(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 $Q=-2×{10}^{-8}C$,

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

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

(d)

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

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

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

(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.