ps03soln - Physics 217 - Homework 3 Solutions Ruxandra...

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Physics 217 - Homework 3 Solutions Ruxandra Bondarescu February 21, 2005 Problem 1. Purcell 2.5 The potential for a basketball size sphere is ϕ = 1000 volts = 3 . 3 statvolts. This potential is the same as the potential of a point charge ϕ = Q/R , where Q is the radius of the sphere and R is the radius of the basketball. The charge per cm 2 is Q 4 πR 2 = ϕ 4 πR = 3 . 3 statvolts 4 π 20 cm (1) where I assumed that the radius of a basketball is R = 20 cm . The number of electrons is then given by Eq. (1) divided by the charge of an electron #electrons cm 2 = 3 . 3 statvolts 4 π (20 cm) (5 10 - 10 esu) = 2 . 62 10 7 (2) So, the number of extra electrons per cm 2 is about 3 × 10 7 . Problem 2. Purcell 2.13 ∇ × E = ˆ z ( ∂E y ∂x - ∂E x ∂y ) = ˆ z (6 x - 6 x ) = 0 (3) which checks that the Feld given in problem 2.1 is a proper electrostatic Feld. ∇ · E = ∂E x ∂x + ∂E y ∂y = 6 y + ( - 6 y ) = 0 (4) This shows that there are no charges enclosed. Problem 3. Purcell 2.18 Method 1. The potential di±erence Δ ϕ = - i center edge E · d l . We discretize the cylinder in rings of charge. The Feld due to an arbitrary ring of charge dq of width dx is a ring of charge on axis is dE = dE x ˆx = xdq ( x 2 + a 2 ) 3 / 2 ˆx (5) 1
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The total electric feld at a point x 0 on axis is the sum over all such rings E ( x 0 ) = - i x 0 0 2 πaσxdx ( x 2 + a 2 ) 3 / 2 + i b - x 0 0 2
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This note was uploaded on 08/07/2008 for the course PHYS 2217 taught by Professor Leclair, a during the Spring '06 term at Cornell University (Engineering School).

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ps03soln - Physics 217 - Homework 3 Solutions Ruxandra...

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