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 penny’s 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 − 5 nC after disassembly. b) If the penny had been given a charge of +5 nC, the dime a charge of − 2 nC, and the nickel a charge of − 1 nC, 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 − 2 nC. This is the charge that the can/lid contraption retains after grounding and disassembly. 3.2. A point charge of 20 nC is located at (4,-1,3), and a uniform line charge of -25 nC/m is lies along the intersection of the planes x = − 4 and z = 6. a) Calculate D at (3,-1,0): The total ﬂux density at the desired point is D (3 , − 1 , 0) = 20 × 10 − 9 4 π (1 + 9) − a x − 3 a z √ 1 + 9 point charge − 25 × 10 − 9 2 π √ 49 + 36 7 a x − 6 a z √ 49 + 36 line charge = − . 38 a x + 0 . 13 a z nC / m 2 b) How much electric ﬂux leaves the surface of a sphere of radius 5, centered at the origin? This will be equivalent to how much charge lies within the sphere. First the point charge is at distance from the origin given by R p = √ 16 + 1 + 9 = 5 . 1, and so it is outside. Second, the nearest point on the line charge to the origin is at distance R = √ 16 + 36 = 7 . 2, and so the entire line charge is also outside the sphere. Answer: zero . c) Repeat part b if the radius of the sphere is 10. First, from part b , the point charge will now lie inside. Second, the length of line charge that lies inside the sphere will be given by 2 y , where y satisfies the equation, 16 + y 2 + 36 = 10. Solve to find y = 6 . 93, or 2 y = 13 . 86. The total charge within the sphere (and the net outward ﬂux) is now Φ = Q encl = [20 − (25 × 13 . 86)] = − 326 nC ....
View Full Document
This note was uploaded on 02/07/2010 for the course EE 190-001-20 taught by Professor Ozcan during the Fall '09 term at UCLA.
- Fall '09