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QUIZ_1_Solutions-1

# QUIZ_1_Solutions-1 - E = kq/r2 for point charges E 1 = k(1...

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QUIZ 1 Solutions 1. For a charged conducting sphere, from the surface of the sphere r = R 0 to ∞, the potential is that of a point charge Q, V = kQ/r. So given V at the surface, Q = R 0 V/k giving Q = 0.25 x 540/9 x 10 9 = 15 x 10 -9 C. Then the surface charge density σ is Q/4πR 0 2 = 15 x 10 -9 /4π x 0.25 2 = 19 x 10 -9 C/m 2 . 2. Now with Q determined calculate V at 1 m from the center of the sphere from V = kQ/r. V = 9x10 9 x 15 x 10 -9 /1 = 135 V with V = 0 at r = ∞. Then from W = qΔV We get W = 1 x 10 -6 C x 135 V = 135 x 10 -6 J. 3. For a dipole, V = kpcosθ/r 2 so p = r 2 V/kcosθ. Here p = 1 2 x 4.5/(9 x 10 9 x cos60 o ). So p = 10 -9 Cm. And from p = qd, d = p/q = 10 -9 /10 -6 = 10 -3 m = 1 mm. 4. All three charges contribute to the electric field at the points in question. So using
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Unformatted text preview: E = kq/r2 for point charges, E 1 = k(1 x 10-6 /1.05 2 – 2 x 10-6 / 1.00 2 + 1 x10-6 /0.95 2 ) = 135.56 N/C E 2 = k(1 x 10-6 /2.05 2 – 2 x 10-6 /2.00 2 + 1 x 10-6 /1.95 2 ) = 8.45 N/C 5. If E is proportional to 1/R n , then E 1 /E 2 = (1/R 1 n )/(1/R 2 n ) = (R 2 /R 1 ) n . As R 2 = 2R 1 this says E 1 /E 2 = 2 n = 16 suggesting n = 4. 6. As E points away from + charge and toward – charge, at the center of the square all four E’s point toward the bottom of the square at a 45 o angle to the vertical. So E total = 4 x (k x 2 x 10-6 /(√2/2) 2 )x cos 45 o = 102 x 10 3 N/C pointing down. 7. However V has no direction so in the kq/r terms the +q’s and the –q’s cancel so V = 0....
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