Electric Potential and Energy
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- Ἀπολλωνία Κουβέλης
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1 Electric Potential and Energy Submitted by: I.D The problem: How much energy is needed to create the following configuration? The solution: Let φ i be the potential at the position of the charge q i as a result of all the other charges. φ i = i j kq j r j (1) Numbering the charges from the left upper corner in the clockwise direction we have ( q 1 φ 1 = q kq a kq a + kq ) (2) 2a ( kq q 2 φ 2 = q a + kq a kq ) (3) 2a q 3 φ 1 = q 1 φ 1 (4) q 4 φ 4 = q 2 φ 2 (5) U = 1 q i φ i = kq2 2 a ( 2 4) U < 0 (6) Then i U < 0 (7) From here we can conclude that the work needed to create this system is negative, therefore we need to invest energy in order to dissolve the system. 1
2 Electric Potential The problem: An infinite 1-D crystal structure is built of point charges along the x-axis with the same distance b between the sequential charges. The sign of the charges changes from each charge to the following one. The absolute value of each charge is q. What is the potential energy U of one charge? Hint: ln(1 + x) = x 1 2 x x3 1 4 x The solution: Let us look at some (say) positive charge at the point P and calculate the potential at this point due to the other charges kq ϕ = 2 r n=1 q( 1) n = 2k = 2kq ( 1) n nb b n n=1 n=1 = 2kq ( b )... = 2kq (1 12 b ) +... The potential energy is = 2kq b (1) (2) ln 2 (3) U = qϕ = 2 ln 2 kq2 b (4) 1
3 א. כדי למצוא את העבודה הדרושה לבניית המערכת יש לחשב את העבודה הדרושה להבאת כל אחד מהמטענים למקומו, לפי הסדר. נניח שקודם מביאים את המטען.2µc בשביל זה לא צריך להשקיע אנרגיה u1 = 0 ולכן לא מתבצעת עבודה. אחר כך מביאים את המטען 4µc למרחק של 1cm מהמטען.2µc הפוטנציאל בנקודה שאליה מביאים אותו הוא: v2 = = = v 2 4πε העבודה הדרושה לכך: 6 u2 = 4 10 v2 = 7.2j אחר כך מביאים את המטען 6µc למרחק של 1cm מהמטען 2µc ולמרחק של 1cm מהמטען.4µc הפוטנציאל בנקודה שאליה מביאים אותו הוא: v3 = + = = v 4πε πε העבודה הדרושה לכך: 6 u3 = 6 10 v3 = 32.4j לכן סה"כ עבודה הדרושה לבניית המערכת היא: U = u + u + u = = 39.6 j total ב. הפוטנציאל בנקודה A שווה לסכום הפוטנציאלים הנוצרים כתוצאה מכל אחד מהמטענים בנפרד: v A = + + = πε πε πε = v 6 7 U = q(va v ) 10 = ( ) = 20.08j ג.
4 1 את הפוטנציאל של מטען נקודתי בראשית ניתן לשב על ידי אינטגרל על השדה של המטען. נגדיר את הפוטנציאל באינסוף להיות אפס, ונקבל: r φ r φ = p q, הפוטנציאל E d r = r kq kq dx = x2 x r = kq r אם נסמן את מיקום המטען השלילי כ 0, ואת מיקום המטען החיובי כ: יהיה: φ r = kq r + kq r 1 p q φ r = kq r + kq r + = kˆr p r 2 kq = r (ˆr 1 1 p) (ˆr p) qr qr kq r ˆr p) qr kq r + kq r kq r + ( 1 + r נקרב עבור מרחק גדול: qr ) ˆr p qr kq ˆr p + 1 q 2 r 2 p 2 ) כדי למצוא את השדה החשמלי, צריך לבצע גראדיאנט של הפוטנציאל. ניתן לעבוד בקוארדינטות קרטזיות או קוארדינטות ספריות. בכל מקרה, למען הנוחות מומלץ להגדיר את כיוון z ככיוון אליו מצביע הדיפול. בכדוריות זה יראה כך: kˆr pẑ kp cos θ φ r = = r 2 r 2 E = φ = kp cos θ 1 ( r = kp 2 φ r = kˆr pẑ r 2 = kpz r 3 2 cos θ E = φ = kp z r = kp 3 r z kpz 1 3 r 3 = kpẑ r + 3kpz r 3 r 5 = k p 3( p r) r + r3 r 5 r 3 ˆr sin θ ) ˆθ r 3 בקרטזיות זה יראה כך: וקיבלנו את אותו השדה כמו זה שקיבלנו מחישוב ישיר.
5 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 are φ 2R = 0, grounded shell φ 3R = φ 4R, conductor Q 3 + Q 4 = 2Q, given (1) The potentials at the shells are φ 2R = kq 2R + kq 2 2R + kq 3 3R + kq 4 4R φ 3R = kq 3R + kq 2 3R + kq 3 3R + kq 4 4R φ 4R = kq 4R + kq 2 4R + kq 3 4R + kq 4 4R Solving the equations above, we get (2) (3) (4) Q R = Q, Q 2R = 1 5 Q, Q 3R = 6 5 Q, Q 4R = 4 5 Q Substituting the charges into the formulas for the potential of hollow spheres gives kq 2R, r < R kq( 1 r 1 2R ), R < r < 2R φ = kq( 6 5r 3 5R ), 2R < r < 3R 1 kq 5 R, 3R < r < 4R 4kQ 5r, r > 4R (5) 1
6 The electric field is the derivative of the potential: 0, r < R kq ˆr, R < r < 2R r 2 E = 6 kq 5 ˆr, 2R < r < 3R r 2 0, 3R < r < 4R 4 kq 5 ˆr, r > 4R r 2 The work needed to bring a charge +q from to the center is U = q φ = q(φ c φ ) = kqq 2R (6) (7) 2
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11 W = F dr = q E dr עבודה מוגדרת כזכור: אם נפרק את השדה לסכום של מספר שדות, ונזכר שתמיד נתן להפוך בסדר בין סכום לאינטגרל, נקבל: W = q E i dr i E i dr = q 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 A ל B העבודה מול הספירה היא: W = R r kq r 2 dr = kq r R r = kq R + kq r = Q ( 1 4πε 0 r 1 ) R השדה של הלוח, לעומת זאת, קבוע, וכיוונו מנוגד לכיוון האינטגרציה, ולכן העבודה היא: R σ W = dr = σ R 0 2ε 0 2ε 0 W = Q 4πε 0 ( 1 r 1 ) σ R R 2ε 0 וסך הכל העבודה היא ועכשיו לעבודה במעבר מ A ל C מול הספירה, החישוב של קודם עדיין בסדר, עם ההבדל שהמרחק גדל קצת: W = Q ( 1 4πε 0 r 1 ) 2R מול המשטח, העבודה נשארת אותו הדבר. ניתן לראות זאת על ידי הקוסינוס שמופיע בדוט, או על ידי תכנון מסלול לא ישיר, מ A ל B ורק אז ל C ולכן סך העבודה: W = Q ( 1 4πε 0 r 1 ) σ R 2R 2ε 0 1
12 Electric Potential and Energy The problem: A solid sphere of the radius R is homogeneously charged with the charge Q and put inside an infinite hollow cylinder. The cylinder inner and outer radii are a and b, R < a < b. The cylinder is homogeneosly charged with the charge density ρ. The center of the sphere is on the cylinder axis. Find the potential in the whole space. Hint: Gauss law and superposition. The solution: First we find the field of the cylinder: 1 dre r r dr = 4πkρ in the cylindrical coordinates (r, ϕ, z). We have E r = 4πk r 0, r < a ρr dr = 2πkρ(r r 0 2 a 2 )/r, a < r < b 2πkρ(b 2 a 2 )/r, r > b The potential is calculated as in the previous problem: r 2πkρ(b 2 a 2 ) ln(r/b), φ c = E r dr = 2πkρ[(b b 2 r 2 )/2 a 2 ln(b/r)], 2πkρ[(b 2 a 2 )/2 a 2 ln(b/a)], r > b a < r < b r < a (1) (2) (3) Attention: Here we cannot put φ = 0 for r since the cylinder is infinite and charges are present at arbitrarily large distances from the coordinate origin: indeed φ when r. We, therefore, arbitrarily chose φ = 0 at r = b. For the sphere we have (in spherical coordinates (R, θ, φ), R 2 = z 2 + r 2 ) { kq/r 2, R > R E R = kqr/r 3, R < R (4) The potential of the sphere { kq/r, φ s = kq/r + kq(r 2 R 2 )/2R 3, The total potential is φ = φ c + φ s. R > R R < R (5) 1
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25 Energy of a disc and a rod Submitted by: I.D 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 > b 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 r 1, r 2 be the positions of charge elements on the disc and the rod, respectively. r 1 = (r cos θ, r sin θ, 0) (1) r 2 = (0, 0, z) (2) r = r 2 r 1 = ( 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 de z = k dq z r 3 dq = σda = σrdrdθ The electric field of the disc is E z = R 2π 0 = 2πkσz 0 R The force acting on the rod is F = z+ b 2 z b 2 kσrdrdθz ((r cos θ) 2 + (r sin θ) 2 + z 2 ) = kσz 3 2 ) ( rdr = 2πkσ 1 0 (r 2 + z 2 ) 3 2 E z ẑλdz = 2πkσλ b + R 2 + R 0 ) z R 2 + z 2 rdr (r 2 + z 2 ) 3 2 The force acting on the disc is the same but in the opposite direction. 2π 0 (4) (5) dθ (6) (7) ( z 2) b 2 ( R 2 + z + b ) 2 ẑ (8) 2 2. What is the flux from the rod that is passing through the disc? The force acting on the disc is F = Eσda = σ E z da The requested flux is Φ = E z da (9) (10) 1
26 Then Φ = F σ = 2πkλ b + R 2 + ( z 2) b 2 ( R 2 + z + b ) 2 (11) 2 3. The mutual energy of the two objects The potential due to the disc on the z-axis is (z > 0) φ(z) = 2πkσ( R 2 + z 2 z) (12) The mutual energy of the disc and the rod is z+ b 2 U = λ φ(z )dz = 2πkσλ z b 2 But writing in a different way Then U = z+ b 2 z b 2 z+ b 2 z b 2 dz ( R 2 + z 2 z ) (13) dz f(z ) (14) F z = du dz = f(z = z b 2 ) f(z = z + b 2 ) (15) ( = 2πkσλ b + R 2 + z 2) b 2 ( R 2 + z + b ) 2 (16) 2 2
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