Download :רוכזכ תרדגומ הדובע Z ~ W =

‫עבודה מוגדרת כזכור‪:‬‬
‫~‬
‫‪~ · dr‬‬
‫‪E‬‬
‫‪Z‬‬
‫‪~ =q‬‬
‫‪F~ · dr‬‬
‫‪Z‬‬
‫= ‪W‬‬
‫אם נפרק את השדה לסכום של מספר שדות‪ ,‬ונזכר שתמיד נתן להפוך בסדר בין סכום לאינטגרל‪ ,‬נקבל‪:‬‬
‫‪Z X‬‬
‫‪XZ‬‬
‫~‬
‫~‬
‫~‬
‫‪~ i · dr‬‬
‫‪W =q‬‬
‫‪Ei · dr = q‬‬
‫‪E‬‬
‫‪i‬‬
‫‪i‬‬
‫ולכן ניתן לחשב עבודה בנפרד עבור כל שדה‪ .‬בבעיה הנוכחית‪ ,‬נחשב את העבודה מול השדה של הספירה‪ ,‬ואת‬
‫העבודה מול הלוח האינסופי בנפרד‪.‬‬
‫במעבר מ ‪ A‬ל ‪ B‬העבודה מול הספירה היא‪:‬‬
‫‪R‬‬
‫‬
‫‬
‫‪Z R‬‬
‫‪Q‬‬
‫‪1‬‬
‫‪1‬‬
‫‪kQ‬‬
‫ ‪kQ‬‬
‫‪kQ kQ‬‬
‫‪+‬‬
‫=‬
‫‪−‬‬
‫‪dr = −‬‬
‫‪=−‬‬
‫= ‪W‬‬
‫‪r2‬‬
‫‪r r‬‬
‫‪R‬‬
‫‪r‬‬
‫‪4πε0 r R‬‬
‫‪r‬‬
‫השדה של הלוח‪ ,‬לעומת זאת‪ ,‬קבוע‪ ,‬וכיוונו מנוגד לכיוון האינטגרציה‪ ,‬ולכן העבודה היא‪:‬‬
‫‪Z R‬‬
‫‪σ‬‬
‫‪σ‬‬
‫‪W =−‬‬
‫‪dr = −‬‬
‫‪R‬‬
‫‪2ε0‬‬
‫‪0 2ε0‬‬
‫וסך הכל העבודה היא‬
‫‪σ‬‬
‫‪R‬‬
‫‪−‬‬
‫‪2ε0‬‬
‫‬
‫‪1‬‬
‫‪1‬‬
‫‪−‬‬
‫‪r R‬‬
‫‬
‫‪Q‬‬
‫= ‪W‬‬
‫‪4πε0‬‬
‫ועכשיו לעבודה במעבר מ ‪ A‬ל ‪C‬‬
‫מול הספירה‪ ,‬החישוב של קודם עדיין בסדר‪ ,‬עם ההבדל שהמרחק גדל קצת‪:‬‬
‫‬
‫‬
‫‪Q‬‬
‫‪1‬‬
‫‪1‬‬
‫= ‪W‬‬
‫√‪−‬‬
‫‪4πε0 r‬‬
‫‪2R‬‬
‫מול המשטח‪ ,‬העבודה נשארת אותו הדבר‪ .‬ניתן לראות זאת על ידי הקוסינוס שמופיע בדוט‪ ,‬או על ידי תכנון‬
‫מסלול לא ישיר‪ ,‬מ ‪ A‬ל ‪ B‬ורק אז ל ‪C‬‬
‫ולכן סך העבודה‪:‬‬
‫‬
‫‬
‫‪1‬‬
‫‪1‬‬
‫‪Q‬‬
‫‪σ‬‬
‫√‪−‬‬
‫‪R‬‬
‫= ‪W‬‬
‫‪−‬‬
‫‪4πε0 r‬‬
‫‪2ε0‬‬
‫‪2R‬‬
‫‪1‬‬
Electric Potential and Energy
The problem:
A conducting sphere of radius R and charge +Q, enclosed by a grounded spherical shell (radius 2R)
and surrounded by a spherical conductor with inner radius of 3R, outer radius of 4R and the total
charge −2Q.
• Find the electric potential and the electric field as a function of r
• Find the work needed to bring a charge q from ∞ to the center of the system
The solution:
The boundary conditions


φ2R = 0,
φ3R = φ4R ,


Q3 + Q4 = −2Q,
are
grounded shell
conductor
given
(1)
The potentials at the shells are
kQ kQ2 kQ3 kQ4
+
+
+
2R
2R
3R
4R
kQ kQ2 kQ3 kQ4
φ3R =
+
+
+
3R
3R
3R
4R
kQ kQ2 kQ3 kQ4
φ4R =
+
+
+
4R
4R
4R
4R
Solving the equations above, we get
(2)
φ2R =
(3)
(4)
1
6
4
Q2R = Q , Q3R = − Q , Q4R = − Q
5
5
5
Substituting the charges into the formulas for the potential of hollow spheres gives
 kQ

, r<R

2R


1
1


kQ( r − 2R ) , R < r < 2R
6
3
φ = kQ( 5r
− 5R
) , 2R < r < 3R


1 kQ

−5 R
, 3R < r < 4R



 4kQ
− 5r
, r > 4R
QR = Q
,
1
(5)
The electric field is the


0
,



kQ


,
 r2 r̂
kQ
6
~
E = 5 r2 r̂
,



0
,



 4 kQ
− 5 r2 r̂ ,
derivative of the potential:
r<R
R < r < 2R
2R < r < 3R
3R < r < 4R
r > 4R
(6)
The work needed to bring a charge +q from ∞ to the center is
U = q∆φ = q(φc − φ∞ ) =
kQq
2R
(7)
2
Energy of a disc and a rod
Submitted by: I.D. 040439358
The problem:
A disc of a radius R is charged uniformly with charge density σ. A rod of a length b is charged
uniformly with charge density λ. The rod is perpendicular to the disc (which is in the x − y plane)
and positioned on the axis of symmetry of the disc. The center of the rod is at z > 2b .
1. Calculate, from the direct integration of the field, the force between the objects.
2. What is the flux from the rod that is passing through the disc?
3. Find the mutual energy of the two objects and derive from it the expression for the force.
The solution:
1. The force between the objects
Let ~r1 , ~r2 be the positions of charge elements on the disc and the rod, respectively.
~r1 = (r cos θ, r sin θ, 0)
(1)
~r2 = (0, 0, z)
(2)
~r = ~r2 − ~r1 = (−r cos θ, −r sin θ, z)
(3)
Because of the symmetry of the problem the force is in the z direction only. The electric field due
to the charge element dq on the disc is
dq z
r3
dq = σdA = σrdrdθ
dEz = k
(4)
(5)
The electric field of the disc is
Z R Z 2π
Z R
Z 2π
rdr
kσrdrdθz
Ez =
= kσz
dθ
3
3
0
0
0 (r 2 + z 2 ) 2 0
((r cos θ)2 + (r sin θ)2 + z 2 ) 2 )
Z R
rdr
z
√
= 2πkσz
=
2πkσ
1
−
3
R2 + z 2
0 (r 2 + z 2 ) 2
(6)
(7)
The force acting on the rod is
F~ =
Z
z+ 2b
z− 2b

Ez ẑλdz = 2πkσλ b +
s
b
R2 + z −
2
s
2
−
b
R2 + z +
2
2

 ẑ
(8)
The force acting on the disc is the same but in the opposite direction.
2. What is the flux from the rod that is passing through the disc?
The force acting on the disc is
Z
Z
~
~
F = Eσda = σ Ez da
The requested flux is
Z
Φ = Ez da
(9)
(10)
1
Then

Φ=
F
= 2πkλ b +
σ
s
R2 + z −
b
2
s
2
−
R2 + z +
b
2
2


(11)
3. The mutual energy of the two objects
The potential due to the disc on the z-axis is (z > 0)
p
φ(z) = 2πkσ( R2 + z 2 − z)
(12)
The mutual energy of the disc and the rod is
Z z+ b
Z z+ b
p
2
2
0
0
U =λ
φ(z )dz = 2πkσλ
dz 0 ( R2 + z 02 − z 0 )
(13)
z− 2b
z− 2b
But writing in a different way
Z z+ b
2
U=
dz 0 f (z 0 )
(14)
z− 2b
Then
dU
b
b
= f (z 0 = z − ) − f (z 0 = z + )
dz 
2
2

s
2 s
2
b
b

= 2πkσλ b + R2 + z −
− R2 + z +
2
2
Fz = −
2
(15)
(16)