Analisis vectorial

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CHAPTER 2
2.1. Four 10nC positive charges are located in the z = 0 plane at the corners of a square 8cm on a side.
A fifth 10nC positive charge is located at a point 8cm distant from the othercharges. Calculate the
magnitude of the total force on this fifth charge for  = 0:
Arrange the charges in the xy plane at locations (4,4), (4,-4), (-4,4), and (-4,-4). Then the fifth charge
will beon the z axis at location z = 4

2, which puts it at 8cm distance from the other four. By
symmetry, the force on the fifth charge will be z-directed, and will be four times the z component offorce produced by each of the four other charges.
F =
4 √
2
× q2
4π0d2 =
4 √
2
× (10−8)2
4π(8.85 × 10−12)(0.08)2 = 4.0 × 10−4 N
2.2. Two point charges of Q1 coulombs each are located at (0,0,1)and (0,0,-1). (a) Determine the locus
of the possible positions of a third charge Q2 where Q2 may be any positive or negative value, such
that the total field E = 0 at (0,1,0):
The total field at(0,1,0) from the two Q1 charges (where both are positive) will be
E1(0, 1, 0) =
2Q1
4π0R2 cos 45◦ ay = Q1
4

2π0
ay
where R =

2. To cancel this field, Q2 must be placed on the y axis atpositions y > 1 if Q2 > 0,
and at positions y < 1 if Q2 < 0. In either case the field from Q2 will be
E2(0, 1, 0) =
−|Q2|
4π0
ay
and the total field is then
Et = E1 + E2 =

Q1
4

2π0−
|Q2|
4π0

= 0
Therefore
√Q1
2
=
|Q2|
(y − 1)2
⇒ y = 1± 21/4

|Q2|
Q1
where the plus sign is used if Q2 > 0, and the minus sign is used if Q2 < 0.
(b) What is the locus if the twooriginal charges are Q1 and −Q1?
In this case the total field at (0,1,0) is E1(0, 1, 0) = −Q1/(4

2π0) az, where the positive Q1 is
located at the positive z (= 1) value. We now need Q2 to liealong the line x = 0, y = 1 in order
to cancel the field from the positive and negative Q1 charges. Assuming Q2 is located at (0, 1, z),
the total field is now
Et = E1 + E2 =
−Q1
4

2π0
az...
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