Homework Ch 21

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Unformatted text preview: Ty + Fg = T cos θ − mg = 0 mg cos θ T= Now let’s look at the horizontal direction Fnet,x = Tx + Fe = −T sin θ + Fe = 0 3 θ l E T Q,m d Fe Fg √ l2 − (l − d)2 2dl − d2 = −mg l−d l−d Fe = T sin θ = mg tan θ = mg √ 2 mg 2ldl−d Fe −d = Q= E E √ 2 −3 )(9.8) (2)(0.12)(0.55)−0.12 (10 0.55−0.12 = 15000 = 5.21 × 10−7 C We know that Fe must be towards the right to balance the force, so the charge must have been positive. 21.35 E = E+Q + E−Q = kQ kQ 4kQax − =− 2 2 (x + a) (x − a) (x + a)2 (x − a)2 The negative sign indicates that the field is to the left. 21.46 Consider a infinitesimal piece dy of the wire, at the position +y, the field it creates at point P is: 1 dQ λdy 1 dE = ˆ= r ˆ r 2 2 + y2 4π 0 r 4π 0 x 4 dy r y θ x P By symmetry, the vertical component of the field will cancel out with the vertical component of the fi...
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This note was uploaded on 09/11/2012 for the course PHYSICS 002 taught by Professor Welsch during the Spring '12 term at University of California, Berkeley.

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