PHYS2255/1415 HOMEWORK 1 Solution (2014 Spring)
1.
(b) Since the vector on y‐axis is a 
,
Since the vector on z ‐xis is a 
,
(c)
(d)
= 
cos θ + 
sinθ
= – 
sinθ+ 
cos θ
= 
,
, thus in cylinder coordinate, we have = 56 
– 6 
= 
2.
We can use the binomial expansion
(a) For y = 0.25 and n = 1.5
(i) The leading order is 1, the relative error is 28.4%, bad estimate.
(ii) 1+1.5*0.25=1.375, the relative error is 1.61%, which is a good estimate.
(iii) 1.375+1.5*0.5*0.25*0.25/2=1.3984375, the relative error is 0.064%, which is an even better estimation.
For y = 0.25 and n = ‐3
(i) The leading order is 1, the relative error is 95.3%, a horrible estimate.
(ii) 1‐3*0.25=0.25, the relative error is 51.17%, still a horrible estimate.
(iii) 0.25+(‐3)*(‐4)/2*0.25*0.25=0.625, the relative error is 22.07%, still a bad estimate.
(b) For y = 0.025 and n = 1.5
(i) The leading order is 1, the relative error is 3.64%, which is a good estimate.
(ii) 1+1.5*0.025=1.0375, the relative error is 0.02%, it is a much better estimate.
(iii) 1.0375+1.5*0.5*0.025*0.025/2=1.037734375, the relative error is 9.3×10‐5%, which is an
excellent estimation.
For y = 0.025 and n = ‐3
(i) The leading order is 1, the relative error is 7.69%, a good estimate.
(ii) 1‐3*0.025=0.925, the relative error is 0.39%, a much better estimate.
(iii) 0.925+(‐3)*(‐4)/2*0.025*0.025=0.92875, the relative error is 0.016%, it is an excellent estimate.
In conclusion, when the value of y is smaller compared to 1, the estimates coming from the binomial approximation method are closer to the exact values.
3.
Obviously, we know that the f(x) approaches zero for very large x, that is, 
or f (x) → 0 as x → ±∞ . However, this is not a complete solution as it does not describe how it approaches zero. Using the binomial expansion
Given y1 = a2 / x2 , y2 = –a2 / x2 , n = -3/ 2 ,
4.
Picture the Problem The diagram shows the locations of point charges q1 and q2 and the point on the
–
x axis at which we are to findE . From symmetry considerations we can conclude that they
–
component of E at any point on the x axis is zero. We can use Coulomb’s law for the electric field due
to point charges and the principle of superposition of fields to find the field at any point on the x axis. We can establish the results called for in Parts (b) and (c) by factoring the radicand and using the
approximation 1 + α ≈ 1 whenever α << 1.
y
q1 = q
a
r
θ
P
0
x
x
θ
E
q1
q2 = q
−a
| (a) Express the x‐component of the electric field due to the point
charges at y = a andy = –a as a function of the distance r from either charge to point P: |
 |
| Substitute for cos θ and r to obtain: | 
|
| –
The magnitude of Ex is: The partial derivative |
|
Setting 
=0 gives the position of greatest electric strength on the x‐axis:
(b) For 
x 
<< a,
x2 + a2 ≈ a2, so:
For 
x 
>> a,
The charges separated by a would appear to be a single charge of
magnitude 2q.
Its field would be given by:
Factor the radicand to obtain:
a2
For a << x, 1 +
and:
≈ 1
x2
(c)
Ex ≈ ( 
3 
=
| 2kqx
3 a |
Ex = 
Ex = 2kqx|L「x2 |((1 + 
,) 
-3 
2
Ex = 2kqx[x2 ]-3i2 =
| 2kq
2 x |
5.
Picture the Problem Let the origin be at the lower left‐hand corner and designate the point charges as shown in the diagram. We can apply Coulomb’s law for point charges to find the forces exerted on q1 by q2, q3, and q4 and superimpose these forces to find the net force exerted on q1. In Part (b), we’ll use Coulomb’s law for the electric field due to a point charge and the superposition of fields to find the electric field at point P(0, L/2).
(a) Using superposition of forces,
express the net force exerted on
q1:
Apply Coulomb’s law to
–
express F2 ,1 :
Apply Coulomb’s law to
–
express F4 ,1 :
Apply Coulomb’s law to
–
express F3,1 :
y
L
q3 = +q
q2 = −q
E2
E1
P(0, L/ 2)
L
E3
L 2 E4
F2,1
q1 = +q
x q4 = −q
F4,1
F3,1
F1 = F2 ,1 + F3,1 + F4 ,1
Substitute and simplify to obtain:
(b) Using superposition of fields, express the resultant field at
point P:
–
Use Coulomb’s law to express E1 :
–
Use Coulomb’s law to express E2 :
–
Use Coulomb’s law to express E3 :
–
Use Coulomb’s law to express E4 :
| –
F1 = = |
kq2
L2 kq2 L2 |
–   + )+ 
( + )- |
+ )=
|(1 – 1 ) 
( + )
( 2 2 ,
kq2
L2
EP = E1 + E2 + E3 + E4 (1)
E–1 = 
1,P = 
), 
E– = kq2 
= k(- q) (|L ) 
2 r22,P 2 ,P r23,P ( 2 ,
E–3 = 
3,P = 
– L – ,) 
Substitute in equation (1) and simplify to obtain:
E–P = 
+ 
+ 
– – ), 
+ 
– ,) 
=
 ((1 –    ,)  |
6.
Picture the Problem The line charge and point (0, y) are shown in the
–
diagram. Also shown is a line element of length dx and the field dE its
–
charge produces at (0, y). We can find dEx from dE and then integrate from x = x1 to x = x2.
Express the x component of
–
dE :
Integrate from x = x1 to x2 and
simplify to obtain:
From the diagram we see that:
and
Substitute to obtain: 
‐ ‐ ‐ ‐ ‐ ‐ THE END ‐ ‐ ‐ ‐ ‐ ‐
评论0