This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Chapter 3 3.1. An empty metal paint can is placed on a marble table, the lid is removed, and both parts are discharged (honorably) by touching them to ground. An insulating nylon thread is glued to the center of the lid, and a penny, a nickel, and a dime are glued to the thread so that they are not touching each other. The penny is given a charge of +5 nC, and the nickel and dime are discharged. The assembly is lowered into the can so that the coins hang clear of all walls, and the lid is secured. The outside of the can is again touched momentarily to ground. The device is carefully disassembled with insulating gloves and tools. a) What charges are found on each of the five metallic pieces? All coins were insulated during the entire procedure, so they will retain their original charges: Penny: +5 nC; nickel: 0 ; dime: 0 . The pennys charge will have induced an equal and opposite negative charge (-5 nC) on the inside wall of the can and lid. This left a charge layer of +5 nC on the outside surface which was neutralized by the ground connection. Therefore, the can retained a net charge of 5nC after disassembly. b) If the penny had been given a charge of +5nC, the dime a charge of 2nC, and the nickel a charge of 1nC, what would the final charge arrangement have been? Again, since the coins are insulated, they retain their original charges. The charge induced on the inside wall of the can and lid is equal to negative the sum of the coin charges, or 2nC. This is the charge that the can/lid contraption retains after grounding and disassembly. 3.2. A point charge of 12 nC is located at the origin. four uniform line charges are located in the x = 0 plane as follows: 80 nC/m at y = 1 and 5 m, 50 nC/m at y = 2 and 4 m. a) Find D at P (0 , 3 , 2): Note that this point lies in the center of a symmetric arrangement of line charges, whose fields will all cancel at that point. Thus D arise from the point charge alone, and will be D = 12 10 9 ( 3 a y + 2 a z ) 4 (3 2 + 2 2 ) 1 . 5 = 6 . 11 10 11 a y + 4 . 07 10 11 a z C / m 2 = 61 . 1 a y + 40 . 7 a z pC / m 2 b) How much electric ux crosses the plane y = 3 and in what direction? The plane intercepts all ux that enters the y half-space, or exactly half the total ux of 12 nC. The answer is thus 6 nC and in the a y direction. c) How much electric ux leaves the surface of a sphere, 4m in radius, centered at C (0 , 3 , 0)? This sphere encloses the point charge, so its ux of 12 nC is included. The line charge contributions are most easily found by translating the whole assembly (sphere and line charges) such that the sphere is centered at the origin, with line charges now at y = 1 and 2. The ux from the line charges will equal the total line charge that lies within the sphere. The length of each of the inner two line charges (at y = 1) will be h 1 = 2 r cos 1 = 2(4)cos sin 1 1 4 = 1 .....
View Full Document
- Fall '08
- Electrical Engineering