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# HW 10 - terry(ect328 – homework 10 – Turner –(59130 1...

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Unformatted text preview: terry (ect328) – homework 10 – Turner – (59130) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An insulating sphere of radius 15 cm has a uniform charge density throughout its vol- ume. 15 cm 5 . 9371 cm 22 . 2702 cm p If the magnitude of the electric field at a distance of 5 . 9371 cm from the center is 54649 N / C, what is the magnitude of the electric field at 22 . 2702 cm from the center? Correct answer: 62637 . 4 N / C. Explanation: Let : R = 15 cm , E 1 = 54649 N / C , r 1 = 5 . 9371 cm , r 2 = 22 . 2702 cm , V = 4 3 π R 3 , and ρ = Q V . R r 1 r 2 p Method 1: We know the magnitude of the electric field at a radius r 1 = 5 . 9371 cm (corresponding to a smaller sphere with sur- face area A 1 and volume V 1 ): the magnitude is E 1 = 54649 N / C. We want to find the magnitude E 2 at a radius r 2 = 22 . 2702 cm, corresponding to a sphere with surface area A 2 and volume V 2 that is larger than the insulating sphere. From Gauss’s Law, we know that since the flux is constant over the sphere, E 1 A 1 = Φ 1 = Q 1 ǫ relating the flux through the Gaussian sphere of radius r 1 to the charge enclosed, Q 1 . We also know Q 1 = ρ V 1 . For the outer sphere (radius r 2 = 22 . 2702 cm), E 2 A 2 = Φ 2 = Q ǫ with Q = ρ V (Not ρ 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 = ρ V ǫ ◦ ρ V 1 ǫ ◦ = 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 π R 2 and V = 4 3 π 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 = (15 cm) 3 (5 . 9371 cm) (22 . 2702 cm) 2 (54649 N / C) = 62637 . 4 N / C . terry (ect328) – homework 10 – Turner – (59130) 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 ρ inside the insulating sphere. Select a spherical Gaussian surface with r = 5 . 9371 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 = ρ V ′ , where ρ is the charge per unit volume and V ′ is the volume enclosed by the Gaussian surface, given by V ′ = 4 3 π r 3 for a sphere. Therefore, q in = ρ V ′ = ρ parenleftBig 4 3 π r 3 parenrightBig ....
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HW 10 - terry(ect328 – homework 10 – Turner –(59130 1...

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