ps01soln

# ps01soln - Physics 217 Homework 1 Solutions Ruxandra...

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Unformatted text preview: Physics 217 - Homework 1 Solutions Ruxandra Bondarescu February 1, 2005 Problem 1. Purcell 1.5 We divide the semicircle in small segments dl where each segment has a charge dq . The electric Figure 1: We discretize the given continuos line charge density λ = dq/dl where dl is the infinitesimal length of arc. The electric field generated by an arbitrary charge dq at the center of the semicircle and its projections on the x and y axis are shown. field dE at the origin of the semicircle for an arbitrary segment of charge dq points radially outward (assuming the semicircle is positively charged). We observe that the x-component of the electric field E x vanishes due to the symmetry of the problem (it can be seen from the figure that for each infinitesimal vector dE x there exists an oppositely oriented vector of the same magnitude and they cancel). Thus the total electric field is E = integraldisplay d E = integraldisplay d E y (1) For an infinitesimal segment of charge dq the y-component of the electric field is | d E y | = | d E | sin θ = dq sin θ R 2 (2) The line charge density λ for a thin semicircle of radius R and total charge Q uniformly distributed can be written as λ = Q πR (3) 1 So, d q = λ dl, where dl is the line element. Eq. (2) becomes | d E y | = dlλ sin θ R 2 (4) Using that dl = Rd θ , we can rewrite the equation above and integrate to calculate the total electric field at the origin | E | = integraldisplay | d E y | = integraldisplay π Rdθλ sin θ R 2 (5) Since R and λ are constant we can pull them out of the integral | E | = λ R integraldisplay π dθ sin θ (6) So, the electric field at the center of the semicircle is | E | = 2 λ R (7) or in terms of the total charge Q the magnitude of the electric field is | E | = 2 Q πR 2 (CGS units) (8) and it points in the y-direction E =- 2 Q πR 2 ˆy (for apositively charged semicircle) (9) Note: I assumed a Cartesian coordinate system centered at the origin with the y-axis is pointing up. Problem 2. Purcell 1.6 Figure 2: (a) The force F A acting on the charge at A is the sum of the force F AB acting on A due to the charge at B and the force F AC acting on A due to the charge at C F A = F AB + F AC (10) 2 Writing Coulomb’s law for each pair of charges F AB = kq A q B r 2 AB ˆ r AB ; F AC = kq A q C r 2 AC ˆ r AC (11) where r AB and r AC are the distances between points A, B and A, C, respectively and are equal to the side of the equilateral triangle r AB = r AC = lscript = 0 . 2 m , q A = 3 × 10- 6 C and q B = q C = 2 × 10- 6 C . The angle between F AB and F AB is 60 degrees because their direction corresponds to the lines between the charges, which are the sides of an equilateral triangle (see Fig. 2). We can use the law of cosines to determine the magnitude of F A | F A | 2 = | F AB | 2 + | F AC | 2 + 2cos π 3 | F AB || F AC | (12) So, | F A | = radicalbigg ( kq...
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ps01soln - Physics 217 Homework 1 Solutions Ruxandra...

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