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CHAPTER (2) Electric Charges, Electric Charge Densities and Electric Field Intensity Charge Configuration a) Point Charge: The concept of the point charge is used when the dimensions of an electric charge distribution are very small compared to the distance to the neighboring charges, i.e. the point charge is occupying a very small physical space b) Distributed Charge The charge may be distributed along a line, among a surface or among a volume. (1) Line Charge: + The line charge density ππ is defined as the charge per unit + + + + length. βπΈ πͺ ππ = πππβπβπ οΏ½ οΏ½ποΏ½ βπ So, ππ = π πΈ π π + ++ ++ + + dl + + + dQ = ΟL dl + οΏ½πͺοΏ½ποΏ½ π πΈ = ππ π π The total charge Q of the line can be determined by π π πΈ = οΏ½ π πΈ = οΏ½ ππ π π π π [πͺ] For uniform line charge, where ππ is constant [πͺ] πΈ = ππ π (2) Surface Charge: U The surface charge density ππ is defined as the charge per unit surface area. βπΈ ππ = πππβπβπ βπ So, π πΈ ππ = π π οΏ½πͺοΏ½ π οΏ½ π οΏ½πͺοΏ½ π οΏ½ π π πΈ = ππ π π The total charge Q of the surface can be determined by πΈ = οΏ½ π πΈ = οΏ½ ππ π π For uniform surface charge, where ππ is constant πΈ = ππ πΊ [πͺ] 3) Volume charge density (ππ ) U βπΈ ππ = πππβπβπ βπ So, ππ = οΏ½πͺοΏ½ π οΏ½ π οΏ½πͺοΏ½ π οΏ½ π π πΈ π π dv dQ = Οv dv π πΈ = ππ π π The total charge Q of the surface can be determined by πΈ = οΏ½ π πΈ = οΏ½ ππ π π For uniform volume charge, where ππ is constant U U πΈ = ππ π½ [πͺ] Example: A uniform spherical volume charge density distribution contains a total charge of 10-8 C, if the radius of the sphere =2x10-2 m. Find Οv . Solution: Q = 10 β8 C , r = 2 * 10 β2 m , V = 4Ο 3 4Ο r = (2 * 10 β2 ) 3 = 8 * 10 β6 m 2 3 3 Q 10 β8 β΄ Οv = = = 2.98 * 10 β4 C m β3 V 4Ο * 8 * 10 β8 3 . Example: U A non-uniform spherical volume charge density π distribution with ππ = π C.m-3. Find the total ππ charge contained in the volume of the sphere of radius a [m] P P Solution: U π πΈ = ππ π π πΈ = οΏ½ π πΈ = οΏ½ ππ π π = οΏ½ ππ π π ππππ½ π π½ π π ππ π π πππ [βπππ π½ ][βπππ π½ ]π πΈ = ππ οΏ½ οΏ½ [π]ππ π π π π πΈ = ππ ππ ππ πͺ Coloumbβs Law: Force Between Two Point Charges: The force between two stationary point charges Q1,Q2 is proportional to the product of the two charges and inversely proportional to the square of the distance R between them. F21 R Q1 aΜ12 F12 Q2 οΏ½πβππ = οΏ½βππ = π πΈπ πΈπ οΏ½ π ππ πΊπ πΉπ οΏ½πΉοΏ½βππ πΈπ πΈπ οΏ½ π ππ πΊπ πΉπ οΏ½πΉοΏ½βππ οΏ½οΏ½πΉοΏ½βππ = βπ οΏ½οΏ½πΉοΏ½βππ unit vector from charge Q 1 to Q 2 Where: π charge. R R R R and, πΊπ = ππβπ πππ = π. πππππβππ π/π Force on Point Charge due to n Point Charges: οΏ½βπ on point charge πΈπ due to n point charges can be Force π determined as R1t Q1 π οΏ½βπ = οΏ½ π π=π πΈπ πΈπ οΏ½ οΏ½βππ [π΅] π ποΏ½πΉ ππ πΊπ πΉππ Qt R2t Rn t Qn Example: U οΏ½β in vacuum on a point Q 1 =10-4 C due to a Find the force π point charge Q 2 =2*10-5 C where Q 1 is centered at point (0,1,2)m and Q 2 at (2,4,5) R R R R P P R R Solution: U F= Q1Q2 4ΟΞ΅ 0 R 2 21 β§ a 21 = Q1Q2 R21 3 4ΟΞ΅ 0 R21 R21 = (0 β 2) xΛ + (1 β 4) yΛ + (2 β 5) zΛ = β2 xΛ β 3 yΛ β 3zΛ β΄ R21 = 4 + 9 + 9 = 22 R R P P aΛ 21 = β΄F = R21 β 2 xΛ β 3 yΛ β 3zΛ = R21 22 (0 β4 β 2 β 10 β5 ) β (β2 xΛ β 3 yΛ β 3zΛ) 4Ο β 8.85 β 10 β12 ( 22 ) 2 β 22 N Electric Field Intensity at a Point due to Point Charge Q: It is a vector force acting on a unit (+ve) charge. The electric field intensity due to a point located at distance R from the charge Q is given by: οΏ½π¬ οΏ½β = οΏ½οΏ½β πΈβ π πΈπΉ πππποΏ½ οΏ½ π = οΏ½ οΏ½β πΉ πππππ (π½/π) ππ πΊπ πΉπ ππ πΊπ πΉπ Electric Field Intensity at a Point due to Point Charges Q1 , Q2 ,β¦.., Qn: If we have a system of charges Q 1 , Q 2 β¦Q n . the total electric field at a point is the vector sum of all fields due to the different charges. οΏ½οΏ½β οΏ½π¬βπ = βππ=π πΈπ πΉ π π οΏ½οΏ½πΉοΏ½βππ [π΅/πͺ] ππ πΊ πΉ π π οΏ½π¬βπ = οΏ½ π=π πΈπ οΏ½ οΏ½βπ [π΅/πͺ] π ποΏ½πΉ ππ πΊπ πΉπ Ξ± R1 Q1 aΜ1 R2 aΜ 2 Q2 Rn aΜ n Qn Example: Find the electric field intensity at the point (2,4,5) m, due to a point charge Q = 2*10-5 C, located at (0,1,2) m. Solution: QR E = 4ΟΞ΅ 0 R 3 R = (2 β 0) xΛ + (4 β 1) yΛ + (5 β 2) zΛ = 2 xΛ + 3 yΛ + 3z R = 4 + 9 + 9 = 22 E= 2 β 10 β5 (2 xΛ + 3 yΛ + 3zΛ) 4Ο (8.85 β 10 )( 22 ) β12 3 = ... xΛ + ... yΛ + ... zΛ Electric Field Intensity at a point p (rc, Ο, z) due to line charge οΏ½οΏ½πΉοΏ½β π πΈ π οΏ½οΏ½β = π π¬ ππ πΊπ πΉπ z dE R οΏ½π¬β = οΏ½ π dl a οΏ½β π πΈ οΏ½πΉ = ππ πΊπ πΉπ π b y x οΏ½β ππ π π οΏ½πΉ ππ πΊπ πΉπ Electric Field Intensity at a Point p (rc, Ο, z) due to Uniform Line Charge Along z-Axis z b dz' R ββ Ξ±2 z' Ξ±1 z a (rc,Ο,z) z y Ο x R rc οΏ½οΏ½β = π π¬ οΏ½οΏ½πΉοΏ½β π πΈ π ππ πΊπ πΉπ π πΈ = ππ π π = ππ π πβ² οΏ½οΏ½β = π οΏ½ππ ππ β π οΏ½π (πβ² β π) πΉ πΉ = οΏ½πππ + (πβ² β π)π οΏ½β = π π¬ note: οΏ½οΏ½β οΏ½ππ ππ β π οΏ½ π (π β² β π ) π πΉ οΏ½πΉ π = οΏ½οΏ½β = πΉ οΏ½πππ + (πβ² β π)π οΏ½ππ ππ β π οΏ½π (πβ² β π)οΏ½ ππ π πβ² οΏ½π ππ πΊπ οΏ½πππ + (πβ² β ποΏ½ π π )π οΏ½ π οΏ½π οΏ½ ππ ππ β π οΏ½ π οΏ½πβ² β ποΏ½οΏ½ ππ οΏ½οΏ½β = π¬ οΏ½ π πβ² π ππ πΊπ π π οΏ½π β² οΏ½ππ π + οΏ½ π β ποΏ½ οΏ½ π οΏ½ π π π ποΏ½ π οΏ½ππ + ππ οΏ½ = π ποΏ½ π ππ οΏ½ππ + ππ οΏ½ π οΏ½ π ππ π ποΏ½ π οΏ½ππ + ππ οΏ½ = βπ οΏ½ππ + ππ οΏ½ ποΏ½ π β‘ π ππ β’ π πβ² οΏ½π¬ οΏ½β = οΏ½ ππ ππ οΏ½ π β’ ποΏ½ ππ πΊπ π π π β’ π β² οΏ½ππ + οΏ½π β ποΏ½ οΏ½ β£ π οΏ½πβ² β ποΏ½ π πβ² οΏ½π οΏ½ β π π οΏ½ π π οΏ½π β² οΏ½ππ π + οΏ½ π β ποΏ½ οΏ½ ππ οΏ½β = οΏ½ π π¬ οΏ½π ππ πΊπ ππ π (πβ² β π) πππ [πππ + (πβ² β π)π ] π π οΏ½π + π ποΏ½ οΏ½ π [πππ + (πβ² β π)π ] ποΏ½ π π οΏ½π¬ οΏ½β = οΏ½π π (π β π) (π β π) ππ β οΏ½ ποΏ½ οΏ½ ππ πΊπ ππ [ππ + (π β π)π ]ποΏ½π [ππ + (π β π)π ]ποΏ½π π π π οΏ½π π ππ ππ + β οΏ½οΏ½ οΏ½ π ππ [ππ + (π β π)π ] οΏ½π [ππ + (π β π)π ]ποΏ½π π π π π οΏ½π π οΏ½π ππ π οΏ½οΏ½β = [πππ πΆπ β ππππΆπ ]οΏ½ π¬ οΏ½ π [πππ πΆπ + πππ πΆπ ] + ππ πΊπ ππ ππ π οΏ½π¬β = Note: ππ ππ πΊπ ππ - For So, οΏ½ππ [πππ πΆπ + πππ πΆπ ] + π οΏ½π [πππ πΆπ β ππππΆπ ]οΏ½ οΏ½π infinite line πΆπ = πΆπ = πππ οΏ½π¬β = π οΏ½ππ Projection point Intersection point οΏ½ππ ππ π ππ ππ πΊπ ππ Electric Field Intensity at a point p (0,0,z) on the axis of a ring charged with uniform ΟL of radius a centered at the origin and positioned in (x-y) plane ZΜ dE (0,0, z ) R a YΛ dΞ¦ dl = adΟ dQ = Ο l dl = Ο l adΟ XΜ οΏ½οΏ½β = π π¬ οΏ½οΏ½πΉοΏ½β π πΈ π ππ πΊπ πΉπ π πΈ = ππ π π = ππ πβ²π π πβ² = ππ ππ πβ² οΏ½οΏ½β = βπ οΏ½ππ πβ²π + π οΏ½π π = βπ οΏ½ππ π + π οΏ½π π πΉ π π π πΉ = οΏ½πβ²π π + π = οΏ½π + π οΏ½β = π π¬ οΏ½π¬β = οΏ½οΏ½β οΏ½π π οΏ½ππ π + π βπ πΉ οΏ½οΏ½πΉοΏ½β = = π πΉ οΏ½ππ + ππ οΏ½ππ π + ππ ππ πβ² οΏ½βπ ππ πΊπ [ππ + οΏ½π ποΏ½ π ποΏ½ π ππ ] οΏ½ ππ π + π οΏ½ π ποΏ½ ππ π ππ οΏ½βπ β² οΏ½π¬ οΏ½β = οΏ½ π π ποΏ½ ππ πΊπ π π π π οΏ½π + π οΏ½ ππ π ππ πΊπ [ππ + π ππ ] οΏ½π ππ ππ β² οΏ½ππ π π π + οΏ½ π οΏ½π π π πβ² οΏ½ οΏ½οΏ½ βπ π π οΏ½ππ is not a constant unit vector and Since, the unit vector π it is a function of πβ² , and since So, οΏ½ π ππππβ² + π οΏ½ π ππππβ² οΏ½ ππ = π π ππ ππ β‘β οΏ½ π οΏ½π ππππ π π π β οΏ½ π οΏ½π ππππβ² π π πβ² β€ ππ π β’ π β₯ π οΏ½π¬β = β’ β₯ ππ π ππ πΊπ [ππ + ππ ] οΏ½π β’ β² β₯ οΏ½π π π π +οΏ½ π β£ β¦ π β² β² οΏ½οΏ½β = π¬ ππ π ππ πΊπ [ππ + οΏ½οΏ½β = π οΏ½π π¬ Example: ππ οΏ½π π π πβ² οΏ½ π ποΏ½ π π ππ ] ππ π π ππΊπ [ππ + ποΏ½ π ππ ] A uniform line charge of infinite extent with ππ = ππ ππͺ/π οΏ½οΏ½β at (6,8,3) m. lies on z-axis. Find π¬ Solution: E= Οl rΛc 2ΟΞ΅ 0 rc rc = x 2 + y 2 = 62 + 82 = 10 20 β 10 β9 rΛc β΄E = = 36rΛc V / m 2Ο (8.85 * 10 β2 )10 10 Electric field intensity of a surface charge: z dE (0,0,z0) Ο R rc y ds x οΏ½οΏ½β = π π¬ οΏ½οΏ½πΉοΏ½β π πΈ π ππ πΊπ πΉπ π πΈ = ππ π πβ² = ππ πβ²π π πβ²π π πβ² οΏ½οΏ½β = βπ οΏ½ππ πβ²π + π οΏ½π π πΉ π πΉ = οΏ½πβ²π π +π S οΏ½β = π π¬ οΏ½β = π¬ οΏ½οΏ½β οΏ½π π οΏ½ππ πβ²π + π βπ πΉ οΏ½οΏ½πΉοΏ½β = = π π πΉ οΏ½πβ²π π +π οΏ½ππ πβ²π ππ πβ²π π πβ²π π πβ² οΏ½βπ β²π π ππ πΊπ οΏ½ππ + β² β² π οΏ½ππ οΏ½π ππ ππ οΏ½ππ οΏ½π πβ² οΏ½β«π οΏ½βπ β« π οΏ½ π ππ πΊπ π π οΏ½πβ²π +ππ οΏ½ ππ οΏ½π ποΏ½ + π ποΏ½ π ππ οΏ½ π + β«π οΏ½πβ²π οΏ½π πβ²π ποΏ½ π π β² π οΏ½ππ +π οΏ½ ππ οΏ½π π π πβ² οΏ½ β«π π οΏ½ππ is not a constant unit vector and Since, the unit vector π it is a function of πβ² , and since So, οΏ½ π ππππβ² + π οΏ½ π ππππβ² οΏ½ ππ = π π π π ππ ππ οΏ½πβ²π οΏ½π πβ²π οΏ½π¬β = οΏ½π ππππβ² β π οΏ½π ππππβ² οΏ½π πβ² οΏ½οΏ½ οΏ½ οΏ½βπ π οΏ½ ππ πΊπ π β² π οΏ½ππ + ππ οΏ½ π π π ππ [πβ²π ]π πβ²π οΏ½π π π πβ² οΏ½ +οΏ½ οΏ½ π ποΏ½ π οΏ½πβ² π + ππ οΏ½ π π π Then, π ππ οΏ½π¬β = οΏ½οΏ½ ππ πΊπ π and since, π β«π πβ²π π πβ²π ποΏ½ π π β² π οΏ½ππ +π οΏ½ [πβ²π ]π πβ²π οΏ½πβ²π π + = οΏ½ ποΏ½ π ππ οΏ½ βπ ποΏ½ π π β² π οΏ½ππ +π οΏ½ So, οΏ½π¬β = π οΏ½π οΏ½π¬ οΏ½β = π οΏ½π ππ οΏ½π π π πβ² οΏ½ οΏ½ π π π οΏ½ = οΏ½ π βπ π [ππ +ππ ] οΏ½π + π π [ππ ] οΏ½π οΏ½ ππ βπ π οΏ½ + οΏ½ ποΏ½ ππΊπ π π π π [π + π ] ππ οΏ½π β ππΊπ π οΏ½ππ + ππ οΏ½ ποΏ½ οΏ½ π For infinite surface π β β Note: οΏ½π¬β = π οΏ½π ππ ππΊπ π οΏ½β = π οΏ½π π For infinite surface the electric field οΏ½π¬ ππΊ π and in direction normal to the surface and out of it. Example: Two infinite uniform sheets of charge πππ πππ πππ located at π = ±π as shown in figure. Find the electric field in all regions. Solution: y -1 Ο x=1 Ο x S1 s2 Region 1 : E1 = Οs 1 2Ξ΅ 0 xΛ , β΄ Et = E1 + E2 = E2 = [ Οs 2 2Ξ΅ 0 xΛ ] 1 Ο + Ο s2 xΛ 2 Ξ΅ 0 s1 Region 2 : E1 = Οs 1 ( β xΛ ) , E2 = 2Ξ΅0 1 xΛ β΄ Et = βΟ +Ο s s 1 2 2Ξ΅0 [ ] Οs 2 2Ξ΅0 ( xΛ ) Region 3 : E1 = β Οs 1 xΛ , E2 = β 2Ξ΅0 1 β΄ Et = β Ο s + Ο s xΛ 1 2 2Ξ΅0 [ ] Οs 2 2Ξ΅0 xΛ
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