Solution :
Length of the plastic rod , L = 1.6 m
Total charge on the plastic rod , Q =
C
The rod is divided into 8 pieces.
a). The length of the 8 pieces is , ![$l=\frac{L}{8}$](https://tex.z-dn.net/?f=%24l%3D%5Cfrac%7BL%7D%7B8%7D%24)
![$=\frac{1.6}{8}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1.6%7D%7B8%7D%24)
= 0.2 m
b). Location of the center of the piece number 5 is given as : 0 m, -0.09375 m, 0 m.
c). The charge q on the piece number 5 is given as
![$q=\frac{Q}{L}\times l$](https://tex.z-dn.net/?f=%24q%3D%5Cfrac%7BQ%7D%7BL%7D%5Ctimes%20l%24)
![$q=\frac{-9 \times 10^{-8}}{1.6}\times0.2$](https://tex.z-dn.net/?f=%24q%3D%5Cfrac%7B-9%20%5Ctimes%2010%5E%7B-8%7D%7D%7B1.6%7D%5Ctimes0.2%24)
=
C
d). WE approximate that piece 5 as a point charge and we need to find out the field at point A(0.7 m, 0, 0) only due to the charge.
We know, the Coulombs force constant, k =
So the X component of the electric field at the point A is given as
![$E_x = 8.99 \times 10^9 \times 1 \times 10^{-8} \ \cos \frac{187.628}{0.70625}$](https://tex.z-dn.net/?f=%24E_x%20%3D%208.99%20%5Ctimes%2010%5E9%20%5Ctimes%201%20%5Ctimes%2010%5E%7B-8%7D%20%5C%20%5Ccos%20%5Cfrac%7B187.628%7D%7B0.70625%7D%24)
= -126.15 N/C
The Y component of the electric field at the point A is
![$E_y = 8.99 \times 10^9 \times 1 \times 10^{-8} \ \sin \frac{187.628}{0.70625}$](https://tex.z-dn.net/?f=%24E_y%20%3D%208.99%20%5Ctimes%2010%5E9%20%5Ctimes%201%20%5Ctimes%2010%5E%7B-8%7D%20%5C%20%5Csin%20%5Cfrac%7B187.628%7D%7B0.70625%7D%24)
= -16.93 N/C
Now since the rod and the point A is in the x - y plane, the z component of the field at point A due to the piece 5 will be zero.
∴ ![$E_z=0$](https://tex.z-dn.net/?f=%24E_z%3D0%24)
Thus, ![$E= $](https://tex.z-dn.net/?f=%24E%3D%20%3C-126.15%2C-16.93%2C0%3E%24)