This preview shows pages 1–3. 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: tovar (jdt436) homework 10 Turner (59070) 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points An insulating sphere of radius 14 cm has a uniform charge density throughout its vol ume. 14 cm 6 . 5013 cm 17 . 1401 cm p If the magnitude of the electric field at a distance of 6 . 5013 cm from the center is 39127 . 1 N / C, what is the magnitude of the electric field at 17 . 1401 cm from the center? Correct answer: 56212 . 8 N / C. Explanation: Let : R = 14 cm , E 1 = 39127 . 1 N / C , r 1 = 6 . 5013 cm , r 2 = 17 . 1401 cm , V = 4 3 R 3 , and = Q V . R r 1 r 2 p Method 1: We know the magnitude of the electric field at a radius r 1 = 6 . 5013 cm (corresponding to a smaller sphere with sur face area A 1 and volume V 1 ): the magnitude is E 1 = 39127 . 1 N / C. We want to find the magnitude E 2 at a radius r 2 = 17 . 1401 cm, corresponding to a sphere with surface area A 2 and volume V 2 that is larger than the insulating sphere. From Gausss Law, we know that since the flux is constant over the sphere, E 1 A 1 = 1 = Q 1 relating the flux through the Gaussian sphere of radius r 1 to the charge enclosed, Q 1 . We also know Q 1 = V 1 . For the outer sphere (radius r 2 = 17 . 1401 cm), E 2 A 2 = 2 = Q with Q = V (Not V 2 , as the Gaussian sur face is larger than the actual physical sphere, and no charge is outside of the sphere.), so E 2 A 2 E 1 A 1 = V V 1 = V V 1 . We know the surface area of a sphere is pro portional to the radius squared, and the vol ume is proportional to the cube of the radius (in particular: A = 4 R 2 and V = 4 3 R 3 ), so E 2 A 2 E 1 A 1 = E 2 r 2 2 E 1 r 2 1 and V V 1 = R 3 r 3 1 . Combining, we get E 2 r 2 2 E 1 r 2 1 = R 3 r 3 1 E 2 = R 3 r 1 r 2 2 E 1 = (14 cm) 3 (6 . 5013 cm) (17 . 1401 cm) 2 (39127 . 1 N / C) = 56212 . 8 N / C . tovar (jdt436) homework 10 Turner (59070) 2 Note that in this solution, we did not actually need to remember the specific formulae for the surface area and volume of a sphere as the constants cancelled. Method 2: First, calculate the charge density inside the insulating sphere. Select a spherical Gaussian surface with r = 6 . 5013 cm, concentric with the charge distribution. To apply Gauss law in this situation, we must know the charge q in within the Gaussian surface of volume V . To calculate q in , we use the fact that q in = V , where is the charge per unit volume and V is the volume enclosed by the Gaussian surface, given by V = 4 3 r 3 for a sphere. Therefore, q in = V = parenleftBig 4 3 r 3 parenrightBig ....
View
Full
Document
This note was uploaded on 12/11/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Charge, Work

Click to edit the document details