Solution 8 - homework 08 KANUNGO ANIL Due 4:00 am Question...

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homework 08 – KANUNGO, ANIL – Due: Sep 20 2007, 4:00 am 1 Question 1, chap 25, sect 2. part 1 of 3 10 points Consider a long, uniformly charged, cylin- drical insulator of radius R and charge density 1 . 4 μ C / m 3 . (The volume of a cylinder with radius r and length is V = π r 2 .) R 2 . 6 cm What is the electric field inside the insulator at a distance 2 . 6 cm from the axis (2 . 6 cm < R )? Correct answer: 2055 . 52 N / C (tolerance ± 1 %). Explanation: Given : ρ = 1 . 4 μ C / m 3 = 1 . 4 × 10 - 6 C / m 3 , r = 2 . 6 cm = 0 . 026 m , and ǫ 0 = 8 . 85419 × 10 - 12 C 2 / N / m 2 . Consider a cylindrical Gaussian surface of radius r and length much less than the length of the insulator so that the compo- nent of the electric field parallel to the axis is negligible. r R The flux leaving the ends of the Gaussian cylinder is negligible, and the only contribu- tion to the flux is from the side of the cylinder. Since the field is perpendicular to this surface, the flux is Φ s = 2 π r ℓ E , and the charge enclosed by the surface is Q enc = π r 2 ℓ ρ . Using Gauss’ law, Φ s = Q enc ǫ 0 2 π r ℓ E = π r 2 ℓ ρ ǫ 0 . Thus E = ρ r 2 ǫ 0 = ( 1 . 4 × 10 - 6 C / m 3 ) (0 . 026 m) 2 (8 . 85419 × 10 - 12 C 2 / N / m 2 ) = 2055 . 52 N / C . Question 2, chap 25, sect 2. part 2 of 3 10 points Determine the absolute value of the po- tential difference between r 1 and R , where r 1 < R . (For r < R the electric field takes the form E = C r , where C is positive.) 1. | V | = C parenleftbigg 1 r 2 1 1 R 2 parenrightbigg 2. | V | = C ( R 2 r 2 1 ) 3. | V | = 1 2 C ( R 2 r 2 1 ) correct 4. | V | = C r 1 5. | V | = C r 1 2 6. | V | = C ( R r 1 ) 7. | V | = C parenleftbigg 1 r 1 1 R parenrightbigg 8. | V | = C ( R r 1 ) r 1 9. | V | = 1 2 C ( R r 1 ) r 1
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homework 08 – KANUNGO, ANIL – Due: Sep 20 2007, 4:00 am 2 10. | V | = C radicalBig R 2 r 2 1 Explanation: The potential difference between a point A inside the cylinder a distance r 1 from the axis to a point B a distance R from the axis is Δ V = integraldisplay B A vector E · vector ds = integraldisplay R r 1 E dr , since E is radial. Δ V = integraldisplay R r 1 C r dr = C r 2 2 vextendsingle vextendsingle vextendsingle vextendsingle R r 1 = C parenleftbigg R 2 2 r 2 1 2 parenrightbigg . The absolute value of the potential differ- ence is | Δ V | = C parenleftbigg R 2 2 r 2 1 2 parenrightbigg = 1 2 C ( R 2 r 2 1 ) . Question 3, chap 25, sect 2. part 3 of 3 10 points What is the relationship between the po- tentials V r 1 and V R ? 1. V r 1 < V R 2. None of these 3. V r 1 > V R correct 4. V r 1 = V R Explanation: Since C > 0 and R > r 1 , from Part 2, V B V A = Δ V = 1 2 C ( R 2 r 2 1 ) < 0 since C > 0 and R > r 1 . Thus V B < V A and the potential is higher at point A where r = r 1 than at point B, where r = R . Intuitive Reasoning: The natural ten- dency for a positive charge is to move from A to B, so A has a higher potential. Question 4, chap 25, sect 3. part 1 of 4 10 points A charge 1 . 1 × 10 - 5 C is fixed on the x - axis at 7 m, and a charge 1 . 1 × 10 - 5 C is fixed on the y -axis at 5 m, as shown.
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