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Unformatted text preview: Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution 2: Electric potential TA: Yi Mao, [email protected] 1. Purcell 1.31: Charged soap bubble. z δ F θ x y φ R Figure 1: A spherical bubble on which charges are uniformly distributed. Refer to figure 1 in which we want to calculate the total force on the upper hemisphere. For any piece of area element dA , the force is pointing outward from the origin; ap plying eq.(35) of Purcell ch.1 (note that “F” in eq.(35) is the force per unit area), the magnitude is determined by δF = 1 2 ( E inner + E outer ) σ dA (1) = Q 2 8 πR 2 sin θdθdφ. (2) where E inner = , (3) E outer = 4 πσ, (4) dA = R 2 sin θdθdφ (5) σ = Q/ (4 πR 2 ) . (6) Eq.(3) is easily seen by eq.(31) of Purcell ch.1. The projected component of δF on XY plane should be completely canceled throughout the hemisphere as σ is constant. So the net force on the hemisphere is the sum of the zcomponent of δF , i.e. F = X δF cos θ = Z π/ 2 dθ Z 2 π dφ Q 2 8 πR 2 sin θ cos θ (7) = Q 2 8 R 2 . (8) 2. Purcell 2.4: ~ E and ρ from φ . For x 2 + y 2 + z 2 < a 2 , φ = x 2 + y 2 + z 2 , ~ E = ~ ∇ φ = ∂φ ∂x ˆ x ∂φ ∂y ˆ y ∂φ ∂z ˆ z = 2 x ˆ x 2 y ˆ y 2 z ˆ z , or ~ E = ( E x ,E y ,E z ) = ( 2 x, 2 y, 2 z ). ρ = 1 4 π ~ ∇· ~ E = 1 4 π ( ∂E x ∂x + ∂E y ∂y + ∂E z ∂z ) = 3 2 π . For x 2 + y 2 + z 2 > a 2 , φ = a 2 + 2 a 3 ( x 2 + y 2 + z 2 ) 1 / 2 , ~ E = ˆ 2 a 3 x ( x 2 + y 2 + z 2 ) 3 / 2 ! ˆ x + ˆ 2 a 3 y ( x 2 + y 2 + z 2 ) 3 / 2 ! ˆ y + ˆ 2 a 3 z ( x 2 + y 2 + z 2 ) 3 / 2 ! ˆ z . ρ = 0 . That’s not the end of the story: we should be careful of the boundary! Take a limit of x 2 + y 2 + z 2 → a 2 from both inside and outside; the fields are not the same. This indicates there are some surface charge density σ...
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This homework help was uploaded on 04/11/2008 for the course PHYSICS 8.022 taught by Professor Hughes during the Spring '08 term at MIT.
 Spring '08
 Hughes
 Charge, Electric Potential, Mass

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