Two observers in different inertial reference frames moving relative to each other at nearly the speed of light see the same two
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(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
(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) The total charge is
. To get the charge on each piece, we should divide this value by 8, the number of pieces:
(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):
If we approximate piece 6 as a single charge, the electric field is given by
where
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
poiting towards the center of piece 6, since the charge is negative.
(e) missing details on this question.
(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) The total charge is
(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):
If we approximate piece 6 as a single charge, the electric field is given by
where
poiting towards the center of piece 6, since the charge is negative.
(e) missing details on this question.