Homework_04-solutions

Homework_04-solutions - toupal(rgt374 – Homework 04 –...

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Unformatted text preview: toupal (rgt374) – Homework 04 – Chiu – (58295) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A variable air capacitor used in tuning circuits is made of N semicircular plates each of radius R and positioned d from each other. A second identical set of plates that is free to rotate is enmeshed with the first set. R θ d Determine the capacitance as a function of the angle of rotation θ , where θ = 0 corre- sponds to the maximum capacitance. 1. C = ǫ (2 N ) d θ R 2 2. C = ǫ N R 2 ( π − θ ) d 3. C = ǫ R 2 ( π − θ ) d 4. C = ǫ N R 2 π d 5. C = ǫ (2 N ) R 2 θ d 6. C = N R 2 ( π − θ ) d 7. C = ǫ N R 2 θ d 8. C = 2 ǫ R 2 (2 π − θ ) d 9. C = ǫ (2 N − 1) R 2 ( π − θ ) d correct 10. C = ǫ N d 2 θ R Explanation: Considering the situation of θ = 0, the two sets of semicircular plates in fact form 2 N − 1 capacitors connected in parallel, with each one having capacitance C = ǫ A d 2 = ǫ π R 2 2 d 2 = ǫ π R 2 d . So the total capacitance would be C = (2 N − 1) ǫ π R 2 d . Note: The common area of the two sets of plates varies linearly when one set is rotating, so the capacitance at angle θ is C = ǫ (2 N − 1) R 2 ( π − θ ) d . 002 10.0 points Consider a parallel plate capacitor system with plate charge Q ( Q > 0) and cross sec- tion A of each plate. Top plate Bottom plate q u q ℓ + Q − Q Plate area is A d E gap Denote the charges on the upper and lower surfaces of the top plate by q u and q ℓ , and the magnitude of the electric field within the gap by E gap . Find the quantities q u , q ℓ and E gap . 1. q u = 0 , q ℓ = Q, E gap = Q 2 ǫ A 2. q u = Q 2 , q ℓ = 3 Q 2 , E gap = Q ǫ A toupal (rgt374) – Homework 04 – Chiu – (58295) 2 3. q u = 0 , q ℓ = Q, E gap = Q ǫ A correct 4. q u = Q, q ℓ = 0 , E gap = Q ǫ A 5. q u = Q 2 , q ℓ = Q 2 , E gap = Q 2 ǫ A 6. q u = Q, q ℓ = 0 , E gap = Q 2 ǫ A 7. q u = − Q 2 , q ℓ = 3 Q 2 , E gap = Q ǫ A 8. q u = − Q, q ℓ = 2 Q, E gap = Q ǫ A 9. q u = − Q, q ℓ = 2 Q, E gap = Q 2 ǫ A 10. q u = Q 2 , q ℓ = Q 2 , E gap = Q ǫ A Explanation: Above the first plate, we can use Gauss’s law and enclose both plates in a Gaussian surface. We find that the enclosed charge is + Q − Q = 0 so the electric field has to be zero. The electric field outside of a conductor is given by E ⊥ = Δ Q encl ǫ Δ A . (1) Since the field is zero above the top plate, the charge on the upper surface must be zero: q u = 0 . Since the net charge on the top plate is + Q , this charge must then all reside on the lower surface of the top plate: q ℓ = Q ....
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Homework_04-solutions - toupal(rgt374 – Homework 04 –...

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