Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
The electric field Submitted by: I.D. 040460479 The problem: An isolated electrical wire is charged uniformly with charge q and bent into a circular shape (radius R) with a small hole b R (where b is the arc length). What is the electrical field in the middle of the circle? The solution: The simple solution is to use superposition. The electric field in the middle of a complete ring is, of course, zero. Now we’ll sum up the field of a complete ring with the field of a very small wire of the same size and shape as the hole but with a negative charge. ~ = kλb x̂ Because b R the wire can be taken as a negative point charge and, therefore, the field is E R2 q (when we take the hole to be on the X axis and λ = 2πR ). It is also possible to calculate the field directly. Taking ~r = (0, 0, 0) and ~r0 = (R cos θ, R sin θ, 0) we have kdq (−R cos θ, −R sin θ, 0) R3 dq = λRdθ Z −α kλ kλRdθ0 ~ (−R cos θ0 , −R sin θ0 , 0) = (2 sin α, 0, 0) E = 3 R R α ~ = dE (1) (2) (3) where ±α are the angles of the edges of the bent wire (or, the upper and lower limits of the hole). Since b R we can approximate tan α ' sin α = b/2 R . Substituting into the expression for the field we obtain ~ = kλb x̂ E R2 (4) 1 2414 נתונה דיסקה מלאה ודקה הנמצאת במישור x − yומרכזה בראשית. רדיוס הדיסקה Rונתון כי צפיפות המטען עליה היא .σ(ϕ) = σ0 cos2 ϕ .1מצאו את השדה לאורך ציר z .2איך יראה השדה בגבול ?z R פתרון סעיף 1 השדה החשמלי מוגדר לפי: ˆ kdq = )~ r ~(E ) · (~r − ~r0 |~r − ~r0 |3 ̂~r = z z )~r0 = rr̂ = r(cos ϕ, sin ϕ, 0 dq = σ(ϕ)rdrdϕ = σ0 cos2 ϕrdrdϕ kσ0 cos2 ϕrdrdϕ )̂· (z ẑ − rr (r2 + z 2 )3/2 ˆ R 2π ˆ = )~ z ~(E 0 0 נחשב את האינטגרל על הזויות ˆ 2π cos2 ϕdϕ = π 0 2π ˆ cos3 ϕdϕ = 0 0 2π ˆ cos2 ϕ sin ϕdϕ = 0 0 קיבלנו כי התרומה תיהיה רק בכיוון z rdr ˆ )· (z (r2 + z 2 )3/2 החלפת משתנים r z R ˆ ~ z ) = πkσ0 z ~(E 0 =x R/z xdx ˆ = πkσ0 sign(z) − √ 1 · ( )z ̂z (1 + x2 )3/2 1 + x2 0 1 R/z ˆ )~ z ) = πkσ0 sign(z ~(E 0 # " 1 ~ z ) = πkσ0 sign(z) 1 − p ~(E ̂z 1 + R2 /z 2 סעיף 2 ניקח את z Rונשתמש בפיתוח טיילור )πkσ0 R2 sign(z ̂z 2z 2 # ≈ ̂z x 2 1 1 + R2 /z 2 ≈1− √1 1+x " ~ E(~z) = πkσ0 sign(z) 1 − p קיבלנו תלות לפי המרחק מציר zבריבוע הדומה למטען נקודתי וכך באינסוף הדיסקה הינה נקודתית 2 :`ed aeyigd .orhnd lr divxbhpi` ici lr dcy aygl epilr Z E= 0 L ~ = E î E kdq kλ1 dx dE = 2 = r (P − x)2 L kλ1 dx kλ1 kλ1 kλ1 − = = 2 (P − x) P −x 0 P −L P jxhvp ,ipyd hend lr lrety gekd z` aygl zpn lr , P mewina dcyd z` mircei epgp`yk eiykr :divxbhpi` zeyrl F~ = F î ~ F~ = Eq kλ1 kλ1 − λ2 dx dF = Edq = x−L L Z 3L F = 2L kλ1 kλ1 − x−L x λ2 dx = kλ1 λ2 (ln(x − L) − ln(x))3L 2L F = kλ1 λ2 ln 1 2L · 2L 4 = kλ1 λ2 ln 3L · L 3