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Homework 04-solutions

# Homework 04-solutions - sanne(as42476 Homework 04 Yao(59110...

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sanne (as42476) – 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 12 . 9 cm 2 and a capacitance of 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 . 0016317 m. Explanation: Let : A = 12 . 9 cm 2 = 0 . 00129 m 2 , C = 7 pF = 7 × 10 12 F , and ǫ 0 = 8 . 85419 × 10 12 C 2 / N · m 2 . C = ǫ 0 A d d = ǫ 0 A C = ( 8 . 85419 × 10 12 C 2 / N · m 2 ) 7 × 10 12 F × ( 0 . 00129 m 2 ) = 0 . 0016317 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 decreases and the electric po- tential remains fixed. 2. The charge and the electric potential in- crease. 3. The charge increases and the electric po- tential remains fixed. 4. The charge remains fixed and the electric potential increases. correct 5. The charge and the electric potential re- main fixed. 6. The charge remains fixed and the electric potential decreases. 7. The charge and the electric potential de- crease. 8. The charge decreases and the electric po- tential increases. 9. The charge increases and the electric po- tential decreases. 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 0 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 ) of the coaxial cable. Assume the length is much greater than the radii of the cylinders ( b ). + Q , b P , r Q , a

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sanne (as42476) – 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 2 π ǫ 0 r 2 . 2. bardbl vector E bardbl = Q 2 π rℓ . 3. None of these 4. bardbl vector E bardbl = Q 2 4 π r 2 . 5. bardbl vector E bardbl = Q 2 2 π ǫ 0 r ℓ . 6. bardbl vector E bardbl = Q 2 π ǫ 0 r ℓ . correct 7. bardbl vector E bardbl = Q 2 2 π rℓ . 8. bardbl vector E bardbl = Q 2 4 π ǫ 0 r 2 . 9. bardbl vector E bardbl = Q 4 π r 2 . Explanation: Construct a Gaussian cylinder at radius r (for a < r < b ). The charge enclosed is the charge on the wire, + Q . Thus, Φ = contintegraldisplay vector E · d vector A = Q ǫ 0 E 2 π r ℓ = Q ǫ 0 ˆ r bardbl vector E bardbl = Q 2 π ǫ 0 r ℓ .
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