Solution :
Length of the plastic rod , L = 1.6 m
Total charge on the plastic rod , Q = [tex]$-9 \times 10^{-8}$[/tex] C
The rod is divided into 8 pieces.
a). The length of the 8 pieces is , [tex]$l=\frac{L}{8}$[/tex]
[tex]$=\frac{1.6}{8}$[/tex]
= 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
[tex]$q=\frac{Q}{L}\times l$[/tex]
[tex]$q=\frac{-9 \times 10^{-8}}{1.6}\times0.2$[/tex]
= [tex]$-1.125 \times 10^{-8}$[/tex] 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 = [tex]$8.99 \times 10^9 \ N.m^2/C^2$[/tex]
So the X component of the electric field at the point A is given as
[tex]$E_x = 8.99 \times 10^9 \times 1 \times 10^{-8} \ \cos \frac{187.628}{0.70625}$[/tex]
= -126.15 N/C
The Y component of the electric field at the point A is
[tex]$E_y = 8.99 \times 10^9 \times 1 \times 10^{-8} \ \sin \frac{187.628}{0.70625}$[/tex]
= -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.
∴ [tex]$E_z=0$[/tex]
Thus, [tex]$E= <-126.15,-16.93,0>$[/tex]
