Phy 132 - Assignment 6
A.
B. 1. . If it's uniformly charged, that means the charge per unit length for the
little piece is the same ratio as for the whole thing:
So, the answer is
2.
C. 1. The charge per unit length for the little piece is the same ratio as for the
whole thing:
where ds is the length of the piece. Since ds = r dθ,
. So, the answer is
2.
D.
Field of one small piece of the rod, such as the one shown:
It points left because we assume the rod to be positive. (If the rod was actually
negative, that would be taken care of by λ being negative.) Field lines point
away from a positive charge.
Since λ means charge/length, λ = dq/dx. (dq =charge of the little piece, dx = its
length.) Therefore, dq = λdx.
Filling in the expression given for λ, dq = (λoxo/x)dx
Put that dq into the expression for dE, then integrate. (Add the fields of all the
little pieces.):
ans.
(Of course, you're not really supposed to treat infinity like it's a number. You're
supposed to say "the limit as x approaches infinity of one over x "….but you
know what I mean.)
E. The field of one little piece, such as the piece in the picture on the question
sheet:
Put in the substitution given for dq, and integrate:
ans.
F.
The question doesn’t ask for this, but if you did evaluate those integrals, you get