PHY303L HW4

# PHY303L HW4 - thwing(jrt2329 – Homework 04 – Yao...

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Unformatted text preview: thwing (jrt2329) – Homework 04 – Yao – (59110) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A parallel-plate capacitor has a plate area of 17 . 6 cm 2 and a capacitance of 9 . 7 pF. The permittivity of a vacuum is 8 . 85419 × 10 − 12 C 2 / N · m 2 . What is the plate separation? Correct answer: 0 . 00160653 m. Explanation: Let : A = 17 . 6 cm 2 = 0 . 00176 m 2 , C = 9 . 7 pF = 9 . 7 × 10 − 12 F , and ǫ = 8 . 85419 × 10 − 12 C 2 / N · m 2 . C = ǫ A d d = ǫ A C = ( 8 . 85419 × 10 − 12 C 2 / N · m 2 ) 9 . 7 × 10 − 12 F × ( . 00176 m 2 ) = . 00160653 m . 002 10.0 points A parallel-plate capacitor is charged by con- necting it to a battery. If the battery is disconnected and the sep- aration between the plates is increased, what will happen to the charge on the capacitor and the electric potential across it? 1. The charge and the electric potential in- crease. 2. The charge and the electric potential re- main fixed. 3. The charge decreases and the electric po- tential increases. 4. The charge and the electric potential de- crease. 5. The charge increases and the electric po- tential decreases. 6. The charge decreases and the electric po- tential remains fixed. 7. The charge remains fixed and the electric potential decreases. 8. The charge remains fixed and the electric potential increases. correct 9. The charge increases and the electric po- tential remains fixed. Explanation: Charge is conserved, so it must remain con- stant since it is stuck on the plates. With the battery disconnected, Q is fixed. C = ǫ A d A larger d makes the fraction smaller, so C is smaller. Thus the new potential V ′ = Q C ′ is larger. 003 (part 1 of 3) 10.0 points Consider a long coaxial arrangement with a cylindrical wire of radius a along the axis of a thin cylindrical shell of radius b . There is a charge of − Q on the inner wire and charge of + Q on the outer shell. The figure below shows a short segment (length ℓ ) of the coaxial cable. Assume the length is much greater than the radii of the cylinders ( ℓ ≫ b ). ℓ + Q , b P , r − Q , a thwing (jrt2329) – Homework 04 – Yao – (59110) 2 The magnitude of the electric field at P where P is at a radius r , between the wire and the shell, is given by 1. bardbl vector E bardbl = Q 4 π r 2 . 2. bardbl vector E bardbl = Q 2 π ǫ r 2 . 3. bardbl vector E bardbl = Q 2 π rℓ . 4. bardbl vector E bardbl = Q 2 4 π r 2 . 5. None of these 6. bardbl vector E bardbl = Q 2 2 π rℓ . 7. bardbl vector E bardbl = Q 2 2 π ǫ r ℓ . 8. bardbl vector E bardbl = Q 2 4 π ǫ r 2 ....
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## This note was uploaded on 11/11/2009 for the course PHY 59110 taught by Professor Yao during the Spring '09 term at University of Texas.

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PHY303L HW4 - thwing(jrt2329 – Homework 04 – Yao...

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