CHAPTER 4
4.1. The value of Eat P(ρ=2,φ=40
◦,z=3)isgiven as E= 100aρ−200aφ+ 300azV/m.
Determine the incremental work required to move a 20 µCcharge a distance of 6 µm:
a) in the direction of aρ: The incremental work is given by dW =−qE·dL, where in this
case, dL=dρ aρ=6×10−6aρ.Thus
dW =−(20 ×10−6C)(100 V/m)(6 ×10−6m) = −12 ×10−9J=−12 nJ
b) in the direction of aφ:Inthis case dL=2dφ aφ=6×10−6aφ, and so
dW =−(20 ×10−6)(−200)(6 ×10−6)=2.4×10−8J=24nJ
c) in the direction of az: Here, dL=dz az=6×10−6az, and so
dW =−(20 ×10−6)(300)(6 ×10−6)=−3.6×10−8J=−36 nJ
d) in the direction of E: Here, dL=6×10−6aE, where
aE=100aρ−200aφ+ 300az
[1002+ 2002+ 3002]1/2=0.267 aρ−0.535 aφ+0.802 az
Thus
dW =−(20 ×10−6)[100aρ−200aφ+ 300az]·[0.267 aρ−0.535 aφ+0.802 az](6 ×10−6)
=−44.9nJ
e) In the direction of G=2ax−3ay+4az:Inthis case, dL=6×10−6aG, where
aG=2ax−3ay+4az
[22+3
2+4
2]1/2=0.371 ax−0.557 ay+0.743 az
So now
dW =−(20 ×10−6)[100aρ−200aφ+ 300az]·[0.371 ax−0.557 ay+0.743 az](6 ×10−6)
=−(20 ×10−6)[37.1(aρ·ax)−55.7(aρ·ay)−74.2(aφ·ax)+111.4(aφ·ay)
+ 222.9] (6 ×10−6)
where, at P,(aρ·ax)=(aφ·ay)=cos(40◦)=0.766, (aρ·ay)=sin(40◦)=0.643, and
(aφ·ax)=−sin(40◦)=−0.643. Substituting these results in
dW =−(20 ×10−6)[28.4−35.8+47.7+85.3+222.9](6 ×10−6)=−41.8nJ
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