Question

A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. 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 a = < 0.7, 0, 0 > m, we divide the rod into 8 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? â„“ = 0.1625 m (b) what is the location of the center of piece number 6? e m. (c) how much charge is on piece number 6? (remember that the charge is negative.) e coulombs (d) approximating piece 6 as a point charge, what is the electric field at location a due only to piece 6? 6 = e n/c (e) to get the net electric field at location a, we would need to calculate due each of the 8 pieces, and add up these contributions. if we did that, which arrow above would best represent the direction of the net electric field at location a?

Answer 1

(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
$L/8=1.3m/8=0.1625 m$

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
$c_6 = 0.0812m+0.1625m=0.2437 m$

(c) The total charge is $Q=-3 \cdot 10^{-8}C$. To get the charge on each piece, we should divide this value by 8, the number of pieces:
$Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C$

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
$q= -3.75\cdot 10^{-9}C$
If we approximate piece 6 as a single  charge, the electric field is given by
$E=k_e \frac{q}{d^2}$
where $k_e=8.99\cdot 10^9Nm^2C^{-2}$ and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
$E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m$
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.