Download Electric dipole and capacitors

Electric dipole and capacitors
Submitted by: I.D. 039409040
The problem:
In the circuit shown on the work page the capacitors are not charged at the beginning. The switch
S is turned left and the first capacitor is charged. The switch is turned right. What are the charges
on all the capacitors?
The solution:
The capacitors C2 , C3 are connected in a series so their charges will be equal q2 = q3 . Let us replace
C3
them with an equivalent capacitor C23 = CC22+C
with the same charge.
3
When S was turned left C1 was charged with q = C1 V0 . The total charge is conserved after the
switch is turned right. We also know that V is equal on C1 and C23 . Let us write the following
equations:
q = q1 + q2 = q1 + q3
q2
q1
V23 = V1 =
=
C23
C1
1
q1
1
+
=
q2
C2 C3
C1
q1 = q − q2 = C1 V0 − q2
1
1
q2
q2
+
= V0 −
C2 C3
C1
1
1
1
+
+
= V0
q2
C1 C2 C3
V0
q2 = q3 = 1
1
1
+
C1
C2 + C3
q1 = C1 V0 − q2
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
1
Electric Current
Submitted by: I.D. 039622568
The problem:
Given the values: ε1 = 1 V, ε2 = 0.5 V, ε3 = 0.6 V, R1 = R2 = 0.5 Ω, R3 = 1 Ω, R4 = 0.4 Ω,
R5 = R6 = 0.6 Ω, R7 = 0.7 Ω
1. Calculate the current flowing through each resistor, and the potential difference between B
and A when the switch is open.
2. Calculate the current flowing through each resistor, and the potential difference between B
and C when the switch is closed.
The solution:
1. Since the switch is open we can disregard the middle branch of the circuit, leaving only a circuit
with resistors and voltage sources in series. By using Ohm’s Law we find:
Vt = ε1 + ε3 = 1 + 0.6 = 1.6 V
(1)
Rt = R1 + R2 + R3 + R6 + R7 = 0.5 + 0.5 + 1 + 0.6 + 0.7 = 3.3 Ω
ε1 + ε3
1.6
Vt
=
=
= 0.485 A
I =
Rt
R1 + R2 + R3 + R6 + R7
3.3
(2)
(3)
With the current moving counter clockwise because of the direction of the voltage sources.
Now in order to calculate the potential difference between B and A, all we have to do is calculate
the voltage between these points:
VBA = ε1 − I(R1 + R2 ) = 1 − I = 0.515 V
(4)
2. Now that the switch is closed we can no longer disregard the middle brance, and we have to use
Kirchhoff’s laws.
We will choose the right node as our junction and assume that I1 comes from above, I2 from below
and I3 flows to the left.
Now, our first path will be clockwise through R1 , ε1 , R2 , R3 , ε2 , R5 , R4 , and our second will be
likewise clockwise through R1 , ε1 , R2 , R3 , R7 , ε3 , R6 . giving us the following equations:
I1 + I2 = I3
(5)
I1 R1 + ε1 + I1 R2 + I1 R3 + ε2 + I3 R5 + I3 R4 = 0
(6)
I1 R1 + ε1 + I1 R2 + I1 R3 − I2 R7 + ε3 − I2 R6 = 0
(7)
1
Using simple math, we found the currents: I1 = −0.6 A, I2 = 0.3 A and I3 = −0.3 A, while the
negative sign signals that the direction in which these currents flow is opposite to the one we chose.
in order to find the potential difference between points B and C we calculate the voltage, keeping
in mind that I3 now has a new direction. thus:
VBC = I3 (R4 + R5 ) = 0.3(0.4 + 0.6) = 0.3 V
2
(8)
‫כוח לורנץ‬
‫‪ .1‬הכוח החשמלי הוא בכיוון 𝑧 ‪,‬כך שעל מנת לבטלו הכוח המגנטי צריך להיות בכיוון‬
‫𝑧‪.−‬‬
‫הכוחות צריכים להיות זהים בגודלם‪ ,‬ולכן‪:‬‬
‫‪.i‬‬
‫𝑚𝐹 = 𝑒𝐹‬
‫הכוח המגנטי מתקבל לפי‪:‬‬
‫‪.ii‬‬
‫𝐵 × 𝑣𝑞 = 𝑚𝐹‬
‫ולכן השדה המגנטי צריך להיות בכיוון 𝑦‪.−‬כי החלקיק נע בכיוון 𝑥‪.‬‬
‫‪.iii‬‬
‫𝑦𝐵𝑣𝑞 = 𝑧𝐸𝑞‬
‫איננו רוצים כוח בכיוון 𝑦 ולכן 𝒛𝑩 חייב להיות אפס‪.‬‬
‫‪- 𝑩𝒙 .‬לעומת זאת יכול להיות בכל ערך‪ ,‬מכיוון שהשדה בכיווןהתנועה לא משפיע כלל‪.‬‬
‫נבחר מטעמי נוחות שגם הוא יתאפס‪ ,‬ונקבל‪:‬‬
‫‪.iv‬‬
‫𝟎‪,‬‬
‫𝒛𝑬‬
‫𝒗‬
‫‪𝑩 = 𝟎,‬‬
‫‪ .2‬מבחינת כיוונים‪ ,‬המטען השלילי הופך הן את כיוון כוח השדה החשמלי‪ ,‬והן את‬
‫כיוון הכוח המגנטי‪ .‬עם זאת‪,‬מסת האלקטרון קטנה פי בערך ‪ 1840‬ממסת‬
‫הפרוטון‪ .‬אם התנע שלהם שווה‪ ,‬אז מהירותו צריכה להיות גדולהבהרבה ‪ ,‬ולכן‬
‫הכוח המגנטי יהיה גדול בהרבה‪ ,‬והאלקטרון לא ינוע בקו ישר‪.‬‬
‫כאשר התיל מרחף הזרם ניצב לשדה‪:‬‬
‫‪µg‬‬
‫‪I‬‬
‫= ‪µg = IB ⇒ B‬‬
‫שימו לב שאלו הם הכוחות ליחידת אורך!‬
‫הכיוון של השדה המגנטי ‪ B‬יהיה פנימה לתוך הדף‪:‬‬
‫כאשר מסובבים את התיל‪:‬‬
‫)‪∑ F = IB sin α − µg = µg sin α − µg = µg (sin α − 1‬‬
‫‪ - α‬זווית בין כיוון הזרם לבין כיוון השדה המגנטי‪.‬‬
‫)‪a = g (sin θ − 1‬‬
‫⇒‬
‫‪∑ F = µa‬‬