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Unformatted text preview: 1) See figure 1 for the geometry of the problem. & = Â¡ / 3; Â¢ = Â¡ / 6 The surface in which the E field hits perpendicularly is called A 1 , and the 2.0 m by 1.0 m surface with a normal to the surface angle of Â¡ / 3 with respect to the E field is called A . Figure 1 Â£ E = 156 Nm 2 /C = EA 1 = EA cos & So, E = Â£ E /A cos & = (156 Nm 2 /C) / [2 m 2 cos( Â¡ / 3)] = 156 N/C 2) Â£ E = q / Â¤ = 9 x 10-6 Â¥ C / Â¤ = 1.0 N m 2 / C 3) The number of elementary charges in -1.0 Â¥ C is: 1.0 Â¥ C / e = 6 x 10 12 4) To do this, we find the E field at the origin due to each charge individually, and then use superposition to find the total E field due to both charges. The angle between the positive x axis and Q 2 is: & = tan-1 (1.6 / 1.2), and the distance from the origin and Q 2 is d 2 = (1.6 m) 2 + (1.2 m) 2 = 4 m 2 So, the magnitude of the E field at the origin due to Q 2 is: E 2 = (1 / 4 Â¡ Â¤ )(2 Â¥ C / d 2 ) Â¦ 4495.9 N/C The direction of E 2 is radially outward, since Q 2 is a positive charge. This means the direction of E 2 is at an angle & with respect to the negative x axis: Figure 2...
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- Spring '07
- Physics, Electric charge, NC, N/C, outer conductor