# Exam1 - smith (js55595) – Exam1 – Sharma – (20715) 1...

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Unformatted text preview: smith (js55595) – Exam1 – Sharma – (20715) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Given : V sphere = 4 π R 3 3 , and A sphere = 4 π R 2 . Consider a sphere, which is an insula- tor, where charge is uniformly distributed throughout. Consider a spherical Gaussian surface with radius R 2 , which is concentric to the sphere with a radius R . R R 2 p Q is the total charge inside the sphere. The total amount of flux flowing through the Gaussian surface is given by 1. Φ = 4 Q ǫ . 2. Φ = Q ǫ . 3. Φ = Q 4 ǫ . 4. Φ = 2 Q ǫ . 5. Φ = Q 8 ǫ . correct 6. Φ = Q 2 ǫ . Explanation: Basic Concept: Gauss’ Law. Solution: For spherical symmetric case, Φ = 4 π r 2 E = Q encl ǫ . Φ = Q encl ǫ = Q ǫ 4 π 3 parenleftbigg R 2 parenrightbigg 3 4 π 3 R 3 = Q 8 ǫ . 002 (part 2 of 2) 10.0 points The magnitude of the electric field bardbl vector E bardbl at R 2 is given by 1. bardbl vector E bardbl = 2 k Q R 2 . 2. bardbl vector E bardbl = k Q 2 R 2 . 3. bardbl vector E bardbl = k Q 2 2 R 2 . 4. bardbl vector E bardbl = 2 k Q 2 R 2 . 5. bardbl vector E bardbl = k Q 2 R 2 . correct 6. bardbl vector E bardbl = k Q R 2 . Explanation: Gauss’s Law gives us 4 π r 2 E = Q encl ǫ = Q ǫ 4 3 π parenleftbigg R 2 parenrightbigg 3 4 3 π R 3 = Q 8 ǫ , E = Q 4 π parenleftbigg R 2 parenrightbigg 2 8 ǫ = Q 4 π ǫ 2 R 2 = k Q 2 R 2 . smith (js55595) – Exam1 – Sharma – (20715)...
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## This note was uploaded on 02/17/2010 for the course PA PHYS taught by Professor Jy during the Spring '10 term at Texas Pan American.

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Exam1 - smith (js55595) – Exam1 – Sharma – (20715) 1...

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