Fa06Soln6

# Fa06Soln6 - Physics 7B Fall 2006 Homework 6 Solutions 21.32...

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Physics 7B Fall, 2006 Homework 6 Solutions 21.32 + Q – 3 Q + 21.37 At point A, from the diagram, we see that the electric fields produced by the charges will have the same magnitude, and the resultant field will be up. We find the angle θ from tan = (0.050 m)/(0.100 m) = 0.500, or = 26.6°. For the magnitudes of the individual fields we have E 1A = E 2A = kQ / r A 2 = (9.0 × 10 9 N · m 2 /C 2 )(7.0 × 10 –6 C)/[(0.100 m) 2 + (0.050 m) 2 ] = 5.04 × 10 6 N/C. From the symmetry, the resultant electric field is E A = 2 E 1A sin = 2(5.04 × 10 6 N/C) sin 26.6° = 4.5 × 10 6 N/C up . For point B we find the angles for the directions of the fields from tan 1 = (0.050 m)/(0.050 m) = 1.00, or 1 = 45.0°. tan 2 = (0.050 m)/(0.150 m) = 0.333, or 2 = 18.4°. For the magnitudes of the individual fields we have E 1B = kQ / r 1B 2 = (9.0 × 10 9 N · m 2 /C 2 )(7.0 × 10 –6 C)/[(0.050 m) 2 + (0.050 m) 2 ] = 1.26 × 10 7 N/C. E 2B = kQ / r 2B 2 = (9.0 × 10 9 N · m 2 /C 2 )(7.0 × 10 –6 C)/[(0.150 m) 2 + (0.050 m) 2 ] = 2.52 × 10 6 N/C. For the components of the resultant field we have E B x = E 1B cos 1 E 2B cos 2 = (1.26 × 10 7 N/C) cos 45.0° – (2.52 × 10 6 N/C) cos 18.4° = 6.52 × 10 6 N/C; E B y = E 1B sin 1 + E 2B sin 2 = (1.26 × 10 7 N/C) sin 45.0° + (2.52 × 10 6 N/C) sin 18.4° = 9.71 × 10 6 N/C. We find the direction from tan B = E B y / E B x = (9.71 × 10 6 N/C)/(6.52 × 10 6 N/C) = 1.49, or 1 = 56.1°. We find the magnitude from E B = E B x /cos B = (6.52 × 10 6 N/C)/cos 56.1° = 1.17 × 10 7 N/C.

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Fa06Soln6 - Physics 7B Fall 2006 Homework 6 Solutions 21.32...

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