8.022pset1solution

8.022pset1solution - Massachusetts Institute of Technology...

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Unformatted text preview: Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution 1: Electric Forces and Electric Fields TA: Yi Mao, [email protected] 1. Purcell 1.3: Charged volleyballs. 1 2 mg 1 F T d=0.5m h=2.5m α α Figure 1: Charged Volleyballs and the free-body diagram Assume both balls have the same mass m = 0 . 3 kg and the same charge q . This identification indicates the geometrical symmetry as shown in fig.1, that is, both balls stretch the strings with the same angel α with respect to the vertical line, and so the electrostatic force F between balls should be in the horizontal direction. Read off the free-body diagram, where F is the electrostatic force and T the tension in each string. We have: F = 1 4 π² q 2 d 2 (1) F = mg tan α (2) tan α = d/ 2 h (3) Eventually we get q = (4 π² mgd 3 / 2 h ) 1 / 2 , (4) or, q = s mgd 3 2 hk , (5) where k = 1 / (4 π² ) = 8 . 988 × 10 9 in SI units. Plug in data, and we get q = 2 . 9 × 10- 6 coulombs. Note that the charges on both balls are of the same sign, since the force between them is repulsive. 2. Purcell 1.5: Electric field of a charged arc. θ d θ Ο θ x y R x y δΕ δΕ δΕ Figure 2: A semicircular rod uniformly charged, and the contribution from an infinitesimal element arc. The total electric field at the center O is the vector summation of all contributions from infinitely many pieces of element arcs on the semicircle. Fig.2 shows the contribution from one element arc that corresponds to an infinitesimal angular change of dθ . The magnitude of its contribution is δE = ρ ( R dθ ) R 2 , (6) where ρ is the linear charge density of the semicircular rod, ρ = Q πR , (7) and R dθ is the length of the element arc. Note that in eq.(6) we’ve taken k = 1 since Q is in esu. One can easily observe the fact that the magnitude δE keeps the same for ALL pieces of element arcs, since the charge is uniformly distributed. Then it is expected that the contributions for x-components of the field from all elements are completely cancelled, while the contributions for y-components are added up, since we can always find a symmetric piece on the other side of a given element arc, which gives an δE x of opposite sign and an δE y of the same sign. Therefore, the total electric field at O is E O = X allpieces δE y = X allpieces δE sin θ = Z π Q πR 2 sin θdθ = 2 Q πR 2 . (8) The direction of ~ E O is along positive y-axis. Aside: one can check E O,x = ∑ δE x ∝ R π cos θdθ = 0....
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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.

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8.022pset1solution - Massachusetts Institute of Technology...

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